Physics Problems Q&A - Recent questions and answers in Physics Problems
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Powered by Question2AnswerAnswered: Unwrapping a rope out of a disk
http://physics.qandaexchange.com/?qa=3511/unwrapping-a-rope-out-of-a-disk&show=3512#a3512
<p><strong>Part a</strong></p>
<p>Answer a.1 is correct. </p>
<p>Answer a.2 is incorrect because force and torque are not the same. There is only one force $F$ on the CM of the disk. This causes linear acceleration $a=F/M$ to the right say. Force $F$ does not pass through the CM of the disk so it creates a torque $\tau=FR$ about the CM, causing rotational acceleration $\alpha=\tau/J$ anti-clockwise, where $J$ is the moment of inertia about the CM. </p>
<p>Anti-clockwise torque $\tau$ does not cancel out the fore $F$ to the right. These are different quantities so they cannot be added, just as mass and length cannot be added. The resultant force on the disk is not zero. Neither is the resultant torque. </p>
<p>Your intuition is incorrect. Your experiment is correct. But as you realise, it is not always like this. Sometimes the experiment is flawed. However, in this case it is your reasoning (intuition), to explain why the experiment failed, which is flawed. </p>
<p>You mention 3 possible explanations for the experiment not giving the result which you expect : (i) the disk is not perfectly cylindrical, (ii) there is friction between the disk and the ground (which could almost be eliminated if the experiment were performed on ice), and (iii) the velocities at opposite points on the rim of the disk are equal and opposite, so the midpoint between them (the CM) must be stationary. </p>
<p>I don't understand how you think explanation (i) might work. Perhaps it is related to (iii). Explanation (ii) work as follows :</p>
<p>The applied force $F$ acts to the right and a friction force $f$ acts to the left. When the experiment is performed on the ground then $f_g<F$. There is a resultant force $F-f_g$ on the disk to the right, so the CM of the disk moves to the right. </p>
<p>Next you argue that if we could make $f=0$ then the CM of the disk should stay where it is. However when we perform the experiment on ice, which causes less friction $f_i<f_g$, then we have $f_i<f_g<F$. The resultant force on the disk $F-f_i$ is still to the right but it is now <strong>greater</strong> than before, not less, because $f_i<f_g$. So reducing or removing friction (eg by performing the experiment on ice) makes the disk <strong>accelerate faster to the right</strong>. It <strong>does not</strong> make the CM accelerate slower or remain at rest. </p>
<p>Your argument in (iii) about the velocities on the rim of the disk cancelling out is also flawed. These two velocities are the same <strong>relative to the CM</strong> of the disk. They must be if the disk remains circular and does not break apart or change its shape. They could be different if the disk is not perfectly circular, as you suggest in explanation (i). Then they could be different distances from the CM. Nevertheless both points would have the same <strong>angular velocity</strong> about the CM.</p>
<p>However, the two points on the rim do not necessarily have the same velocities <strong>relative to the ground</strong>. So these two velocities do not have to be equal and opposite, and their vector sum (which gives the velocity of the CM) does not have to be zero. The disk does not have to remain stationary.</p>
<p><strong>Parts b and c</strong></p>
<p>I shall deal with these later.</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3511/unwrapping-a-rope-out-of-a-disk&show=3512#a3512Fri, 06 Sep 2019 11:42:58 +0000Answered: Electric Field for the circular path of positively charged particle
http://physics.qandaexchange.com/?qa=3509/electric-field-circular-path-positively-charged-particle&show=3510#a3510
<p>Answer A is correct. </p>
<p>Imagine a fixed negatively charged particle at the focus of the electric field lines. The electric force is a central force. The free positively charged particle orbits in a circle around the fixed negative charge. There are other electric field lines apart from those shown, and only part of the circular path is shown.</p>
<p>A circular path is not the only possibility in a central force field. Parabolic and hyperbolic paths are also possible, as well as straight-line paths heading directly towards or away from the fixed charge. These paths look very nearly circular at the vertex, but further away they get straighter and straighter. The whole path is not circular.</p>
<p>Answer B would be possible if there is another force acting on the positively charged particle, like a friction or viscous force, which almost balances the electric force. The particle would then move at very slow speed, almost following the field lines but overshooting them by a very small amount. The overshoot can be made as small as you wish by making the electric and viscous forces more nearly equal. This kind of a curve is called a <a rel="nofollow" href="https://en.wikipedia.org/wiki/Tractrix"><strong>tractrix</strong></a>.</p>
<p>However, we are told nothing about any other forces so we have to assume that they do not exist.</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3509/electric-field-circular-path-positively-charged-particle&show=3510#a3510Sat, 31 Aug 2019 19:54:26 +0000Stiffness constant $k$ for a diatomic molecule
http://physics.qandaexchange.com/?qa=3508/stiffness-constant-%24k%24-for-a-diatomic-molecule
<blockquote><p>A hypothetical diatomic molecule has a binding length of $0.8860 nm$. When the molecule makes a rotational transition from $l = 2$ to the next lower energy level, a photon is released with $\lambda_r = 1403 \mu m$. At a vibration transition to a lower energy state, a photon is released with $\lambda_v = 4.844 \mu m$. Determine the spring constant k.</p>
</blockquote>
<p>What I've done is:</p>
<p>1) Calculate the moment of inertia of the molecule by equating Planck's equation to the transition rotational energy:</p>
<p>$$E = \frac{hc}{\lambda} = \frac{2 \hbar^2}{I}$$</p>
<p>Thus solving for $I$:</p>
<p>$$I = 1.562 \times 10^{-46} kg m^2$$</p>
<p>2) Knowing that the moment of inertia of a diatomic molecule rotating about its CM can be expressed as $I = \rho r^2$, solve for $\rho$ (where $\rho$ is the reduced mass and $r$ is the distance from one of the molecules to the axis of rotation; thus $r$ is half the bond length. EDIT: $r$ is the bond length and NOT half of it. Curiously, if you use the half value you get a more reasonable k: around $3 N/m$):</p>
<p>$$\rho = \frac{I}{r^2} = 1.99 \times 10^{-28} kg$$</p>
<p>3) Solve for $k$ from the frequency of vibration equation:</p>
<p>$$f = \frac{1}{2\pi}\sqrt{\frac{k}{\rho}}$$</p>
<p>Knowing:</p>
<p>$$\omega = 2\pi f$$</p>
<p>We get:</p>
<p>$$k = \omega^2 \rho = (c/\lambda_v)^2 \rho = 0.763 N/m$$</p>
<p>Where $\lambda_v= 4.844 \times 10^{-6} m$</p>
<p>The problem I see is that the method seems to be OK but the result does not convince me. We know that the stiffness constant $k$ is is a measure of the resistance offered by a body to deformation. The unknown diatomic molecule we're dealing with seems to be much more elastic than $H_2$ (which has $k = 550 N/m$).</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3508/stiffness-constant-%24k%24-for-a-diatomic-moleculeTue, 20 Aug 2019 10:57:21 +0000Answered: Showing rotational energy for a nucleus
http://physics.qandaexchange.com/?qa=3505/showing-rotational-energy-for-a-nucleus&show=3507#a3507
<p>My method is correct and there's a flaw in the exercise; the experimental value for the first excited state of 8-Be nucleus is around $3 MeV$. </p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3505/showing-rotational-energy-for-a-nucleus&show=3507#a3507Sat, 17 Aug 2019 09:18:04 +0000Answered: Theoretical limit of resolution for an electron microscope | Giancoli, Chapter 37 Ex. 43
http://physics.qandaexchange.com/?qa=3495/theoretical-resolution-electron-microscope-giancoli-chapter&show=3501#a3501
<p>Your alternative method uses the non-relativistic approximation twice : first when you insert a value for $v/c$ in the numerator, second when you insert a value for $v$ in the denominator. Because of this your error is compounded, so you get a result which is less accurate than if you inserted the approximation only once.</p>
<p>Whenever you make approximations, it is most accurate to do so only once, at the last possible step, as follows :</p>
<ol>
<li><p>First identify a dimensionless variable $y$ which depends on the independent variable in your problem. Examples : $\beta=v/c$ if the independent variable is $v$ or $x=T/mc^2$ if it is $T=eV$( see below). </p>
</li>
<li><p>Obtain a formula for the quantity you are interested in, as a function of $y$.</p>
</li>
<li><p>Expand the formula into a power series in $y$, using Taylor's Theorem. Check that the expansion is valid for the range of values of $y$ which interest you. For example, the expansion $$(1+y)^n=1+ny+\frac12 n(n-1)y^2+\frac16n(n-1)(n-2)y^3+...$$ is valid provided that$y\lt 1$. </p>
</li>
<li><p>Finally throw out higher powers of the dimensionless variable, depending on how much accuracy you want to have. </p>
</li>
</ol>
<hr>
<p>The accurate formula which you got can be written as $$\lambda=\frac{hc}{\sqrt{2Tmc^2+T^2}}=\lambda_0 (1+\frac12 x)^{-1/2}= \lambda_0 (1-\frac14 x+\frac{3}{32}x^2-\frac{5}{128}x^3-...)$$ where $\lambda_0=\frac{h}{\sqrt{2mT}}$ is the de Broglie wavelength assuming the electron is non-relativistic, and $x=\frac{T}{mc^2}=\frac{eV}{mc^2}$. This expansion is valid because $T=85keV$ and $mc^2=511keV$ (the rest mass of the electron) so $\frac12 x\lt 1$.</p>
<p>Note that the approximation is introduced only at the last step, by ignoring the higher powers of $x$. The full infinite series expansion, and the steps before it, are 100% accurate.</p>
<p>Using the figures given, we get $\lambda_0=4.2066pm, x=0.08317$ and $\lambda\approx4.0317pm$ when we ignore terms in $x^2$ or higher powers. Using the exact equation we get $\lambda=4.0419pm$.</p>
<p>Your method first makes the classical approximation $T=eV\approx \frac12 mv^2$ before any. Then $$\frac{v^2}{c^2}\approx\frac{2T}{mc^2}=4x$$ $$mv\approx\sqrt{2mT}$$ $$\lambda=\frac{h\sqrt{1-\frac{v^2}{c^2}}}{mv}\approx \lambda_0 (1-4x)^{+1/2}\approx \lambda_0 (1-2x-2x^2+4x^3-...)$$ </p>
<p>Compare this power series with the one above. The term in $x$ now has a coefficient of$2$ instead of $\frac14$, so the correction to $\lambda_0$ is $8\times$ what it should be. The previous line was already an approximation, so the power series is also an approximation, regardless of however many terms we retain. Unlike the series above, which gets close to being 100% accurate as we retain more and more powers.</p>
<p>Ignoring terms in $x^2$ or higher powers. we get $\lambda \approx 3.5069pm$ instead of$\lambda \approx 4.0317pm$. </p>
<p>The larger is the value of $T=eV$ then the larger is $x$ also, and the greater is the difference between the correct approximation $\lambda\approx \lambda_0 (1-\frac14 x)$ and your incorrect approximation $\lambda\approx \lambda_0 (1-2x)$. And as noted above, if $x\gt \frac14$ then your approximation $\lambda\approx \lambda_0 (1-4x)^{+1/2}$ gives an imaginary result, whereas the exact formula $\lambda= \lambda_0 (1+\frac12x)^{-1/2}$ gives a real result because $x\ge 0$ for all values of $V$.</p>
<hr>
<p>There is no sharp <strong>cut off point</strong> for $V$ at which the electron suddenly switches over from being classical to relativistic, or above which your approximation suddenly ceases to be a good one. The change is gradual. </p>
<p>At $V=33kV$ the value of $x$ is less than half that at $V=85kV$ so the smaller is the difference between your erroneous calculation and the correct one. At both values of $V$ your method gives an error which is $8\times$ bigger than it should be if you used the correct approximate formula. </p>
<p>In fact, if the accelerating potential is such that $T=eV \gt \frac12 mc^2$ then $v \gt c$ so the factor $\sqrt{1-\frac{v^2}{c^2}}$ in the numerator would be imaginary, and so would the de Broglie wavelength.</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3495/theoretical-resolution-electron-microscope-giancoli-chapter&show=3501#a3501Tue, 13 Aug 2019 14:15:44 +0000Answered: Carnot engine running backwards (heat pump)
http://physics.qandaexchange.com/?qa=3498/carnot-engine-running-backwards-heat-pump&show=3500#a3500
<p>For both machines the COP is defined as the ratio of <strong>useful</strong>energy output to the total energy input. The useful energy outputs of the heat pump and the refrigerator are different, so their coefficients of performance are different. </p>
<p>What is <strong>useful</strong> depends on the purpose of the machine, or what <strong>good performance</strong> means. This is a matter of human choice rather than of thermodynamics. So although the COP is defined in the same way, the result is different because what is useful is different.</p>
<p>The purpose of the heat pump is to dump as much heat as possible at the hot sink. So the useful output of a heat pump is the <strong>total heat which is added to the hot sink</strong> : $Q_h=Q_c+W$. It does not matter whether this heat comes from the cold source or from the engine or motor which makes the transfer happen. So the COP of a heat pump is $$COP_{hp}=\frac{Q_h}{W}$$ </p>
<p>The purpose of a refrigerator is to remove as much heat as possible from the cold source. So the useful output of a refrigerator or air conditioner is the <strong>net heat which is removed from the cold source</strong> : $Q_c$. The energy supplied to the refrigerator motor is not dumped in the cold source, so it does not affect the useful energy output of the refrigerator. The COP of a refrigerator is therefore $$COP_{ref}=\frac{Q_c}{W}=\frac{Q_h-W}{W}=COP_{hp}-1$$</p>
<p>This is the relation which you have found. Your first result is the COP for a refrigerator; the second is the COP for a heat pump.</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3498/carnot-engine-running-backwards-heat-pump&show=3500#a3500Tue, 13 Aug 2019 01:54:25 +0000Answered: Laplace Equation boundary condition - Why are the current densities equal 0?
http://physics.qandaexchange.com/?qa=3488/laplace-equation-boundary-condition-current-densities-equal&show=3497#a3497
<p>There ought to be a diagram showing how the angle $\theta$ is defined. Nevertheless, the boundary condition which you have been given is confusing.</p>
<p>The diagram above shows a vertical plane through the centre of the spherical electrode. There is <strong>cylindrical symmetry</strong> here, so a <strong>cylindrical co-ordinate system</strong> is the obvious choice. If the plane is rotated through any azimuthal angle $\theta$ about the vertical axis $z$ through the centre of the sphere, then all measurements (potentials, current densities, etc) would be the same at all points with the same cylindrical co-ordinates $\rho, z$. So there could be no difference in potential or current density around any circle of radius $\rho$ which is centred on and perpendicular to the $z$ axis.</p>
<p>Inside the conducting electrode there is no resistance (infinite conductivity) so there could be a current of constant density around such a circle inside the electrode, without there being any change in potential around any azimuthal circle : $$J_{\theta}=\text{constant}$$</p>
<p>Inside the regions of finite conductivity 1 & 2, there could not be any current around an azimuthal circle, because there would have to be a change in potential between the start and the end points, which are the same, so this is impossible :$$J_{1\theta}=J_{2\theta}=0$$ </p>
<p>In all three regions the boundary condition applies for <strong>all</strong> values of $\theta$ and particular values of $\rho, z$. There is nothing special about $\theta=\frac12 \pi$. The boundary condition given in the book is confusing. It suggests that there is something special about the azimuthal angle $\theta=\frac12\pi$ but there is nothing in the problem which supports this.</p>
<p>Perhaps a spherical polar system is being used. If so, and if $\phi$ is the polar angle, then $\phi=\frac12 \pi$ defines the horizontal plane through the centre of the sphere. The above boundary condition is still true for this value of $\phi$ and all values of $r$ : $$J_{1\theta}(\phi=\frac12 \pi)=J_{2\theta}(\phi=\frac12 \pi)=0$$ But there is nothing special about this plane. The boundary condition applies for all other planes perpendicular to the axis, and all values of $\phi$ and $r$ provided that $\rho=r\sin\phi$ and $z=r\cos\phi$ are constant.</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3488/laplace-equation-boundary-condition-current-densities-equal&show=3497#a3497Tue, 30 Jul 2019 21:44:08 +0000Answered: Using symmetry, to solve more complex circuits.
http://physics.qandaexchange.com/?qa=3489/using-symmetry-to-solve-more-complex-circuits&show=3494#a3494
<p>By symmetry the current in branch CO equals the current in branch OB. Therefore the two branches CO and OB can be detached from O, while remaining connected to each other.</p>
<p>Branch CB then consists of two parallel branches, one of $r$ and one of $2r$. This combination is equivalent to $\frac23r$. The resistance across ACBB' (where B' is the output node) is then $r+\frac23r+r=\frac83r$.</p>
<p>The same trick can be applied to symmetrical branch ACDB' to reduce it also to $\frac83r$.</p>
<p>The network between input and output nodes is made up of 3 branches in parallel (ACBB', AOB', AEDB') with resistances $\frac83r, 2r, \frac83r$ respectively. The two resistances of $\frac83r$ in parallel are equivalent to $\frac43r$. This combination is in parallel with $2r$ so the total resistance is $$\frac{\frac43r \times 2r}{\frac43r+2r}=\frac45r$$ </p>
<p><img src="https://i.imgur.com/OOMFv3c.png" alt=""></p>
<p>See <a rel="nofollow" href="https://www.stemez.com/subjects/joint_entrance_examination/1rjeephysics/1rjeephysics/1RJeePhysics/1R12-1713.htm">Stemez JEE physics</a>.</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3489/using-symmetry-to-solve-more-complex-circuits&show=3494#a3494Wed, 24 Jul 2019 18:50:08 +0000Answered: Find the angle between the $x$-axis and a vector
http://physics.qandaexchange.com/?qa=3105/find-the-angle-between-the-%24x%24-axis-and-a-vector&show=3492#a3492
<p> You have to subtract 58 deg. from 180 deg. because question is demanding angle from positive x axis while formula give angle from negative x axis. Moreover as sammy said in the comment vector x is pointing in positive direction so you should drop that negative from the answer.</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3105/find-the-angle-between-the-%24x%24-axis-and-a-vector&show=3492#a3492Wed, 24 Jul 2019 13:42:02 +0000graphing height and speed of projectiles
http://physics.qandaexchange.com/?qa=3485/graphing-height-and-speed-of-projectiles
<p>The football ball $m=1kg$ was kicked up vertically with speed $v=30m/s$. Make the model of<br>
movement track, also the graph of speed, knowing that $F_r=bv^2$ and $b=1 Ns^2/m^2$.</p>
<ol>
<li>The football ball $m=1 kg$ was kicked up vertically with speed $v=20 m/s$. Make the graph of speed and the y position.</li>
<li>The stone was thrown horizontally with speed $v=10m/s$ from the building at height $h=15m$. Make the model of movement track, also the graph of speed.</li>
<li>The mass $m=1 kg$ is moving on inclined plane, inclined with an angle $\theta=30$°. Kinetic friction coefficient $μ_k=0.2$. Make the graph of speed and the movement.</li>
</ol>
Physics Problemshttp://physics.qandaexchange.com/?qa=3485/graphing-height-and-speed-of-projectilesWed, 19 Jun 2019 11:05:15 +0000Answered: How to use symmetry?
http://physics.qandaexchange.com/?qa=3473/how-to-use-symmetry&show=3482#a3482
<p>Imagine that the three $4\Omega$ resistors are removed, leaving two separate circuits. </p>
<p>Now join EF with a conducting wire so that they are at the same potential. No current will flow from one circuit into the other, because there is no path for it to return along, and charge which flows cannot accumulate in either of the two circuits. </p>
<p>How does the potential at C compare with that at D? How does the potential at A compare with that at B? The PD across each battery is divided equally across the resistors in the same circuit. Relative to the potentials at E=F as 0V, the potentials at A, B will be 2V and the potentials at C, D will be 1V.</p>
<p>What effect will it have if you re-insert resistors between AB, CD and EF? Since AB, CD and EF are all at the same potential, it will not make any difference if we connect AB, CD with wires or re-insert the $4\Omega$ resistors. No current will flow along these connecting wires/resistors. </p>
<p>So the currents in the $6\Omega$ and $3\Omega$ resistors are the same whether or not the $4\Omega$ resistors are there.</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3473/how-to-use-symmetry&show=3482#a3482Wed, 22 May 2019 16:46:45 +0000Answered: The photocurrent in the detector assuming all rays to be paraxial w.r.t. lens
http://physics.qandaexchange.com/?qa=3472/the-photocurrent-the-detector-assuming-rays-paraxial-lens&show=3481#a3481
<p>The mistake in your calculation is that you are assuming that the light which is focussed by the lens into a point $S'$ at $60cm$ to the right of the lens acts like a <strong>point source</strong> which radiates equally in all directions.</p>
<p>This assumption is not correct. All of the light from the lens converges onto $S'$ in a cone, and diverges again <strong>only to the right</strong> into another cone with the same angle. In the paraxial approximation we can assume this light is spread uniformly across the base of the cone.</p>
<p>When the cone of light which diverges from $S'$ reaches the detector its radius is half that of the aperture of the lens. This can be found from the geometry of similar triangles. So the area $A$ of the circular base of this cone at the detector is $\frac14$ of that of the lens, ie $A=1cm^2$. </p>
<p>The area of the detector is only $0.5cm^2$ and is completely covered by the cone. So only half the light which is collected by the lens subsequently falls onto the detector.</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3472/the-photocurrent-the-detector-assuming-rays-paraxial-lens&show=3481#a3481Wed, 22 May 2019 15:46:34 +0000Answered: Current through a particular branch of RC circuits
http://physics.qandaexchange.com/?qa=3474/current-through-a-particular-branch-of-rc-circuits&show=3476#a3476
<p>Initially no current flows through the $600k\Omega$ resistor. It all flows onto the uncharged capacitor. After a long time the capacitor will be fully charged and the current through the $600k\Omega$ resistor is the same as that through the battery, which is $\frac{80}{3}\mu A$. </p>
<p>So the current through the resistor increases exponentially from $0$ to a limit of $$I_{\infty} = \frac{80}{3} \mu A$$ The rate of increase is set by the <strong>time constant</strong> of the parallel combination of resistor and capacitor, which is $\tau=CR$, where $R=600k\Omega$ and $C=2.5nF$. Such an increase can be written as $$I(t)=I_{\infty}(1-e^{-t/ \tau})$$ </p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3474/current-through-a-particular-branch-of-rc-circuits&show=3476#a3476Wed, 22 May 2019 11:06:56 +0000Answered: Velocity wrt COM
http://physics.qandaexchange.com/?qa=3465/velocity-wrt-com&show=3470#a3470
<p>Both spheres have their initial velocities reversed when they come into contact again. This is a consequence of both the <strong>Conservation of Momentum</strong> and the <strong>Conservation of Energy</strong>.</p>
<p>This is essentially the same as throwing a ball up into the air. If air resistance can be ignored, so that there is no loss of energy, it returns to the Earth with the same speed as it was launched. The Earth also returns to the ball with the same speed at which it initially moved away.</p>
<p>In this case also there is no loss of energy between the initial impulse and the final collision between the spheres. The final gravitational potential energy is the same as the initial potential energy, because the spheres return to their initial positions. So the final total kinetic energy will be the same as the initial total kinetic energy.</p>
<p>In the COM frame the momentum of the big sphere is always equal and opposite to that of the small sphere. (The centre of mass frame is the same as the centre of momentum frame.) Therefore if there are no external forces, so that momentum is conserved, then the speed of the small sphere is always in the same proportion $\frac{M}{m}$ to that of the big sphere, and the KE of the small sphere is also always in the same proportion.</p>
<p>When this consequence of the conservation of momentum is combined with the conservation of energy, it can be deduced that the speeds of both particles will be the same when they return to their initial positions. </p>
<p>If there was a loss of energy due to internal forces the two particles would not have their initial speeds when they returned to the initial position, but because the conservation of momentum still applies the ratio of their two speeds would remain at all times the same as initially, and they would return to the initial position at exactly the same time.</p>
<p>If the ratio of velocities varied, while the total energy was conserved, then one particle would return to the starting point before the other. This would mean that the position of the COM had changed, which cannot happen if there are no external forces.</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3465/velocity-wrt-com&show=3470#a3470Thu, 16 May 2019 19:54:06 +0000Answered: Net energy loss in infinite series of capacitors
http://physics.qandaexchange.com/?qa=3460/net-energy-loss-in-infinite-series-of-capacitors&show=3469#a3469
<p>You have forgotten that there is work done by the battery after $t=0$, so that there is energy lost before S1 is opened and S2 is closed.</p>
<p>The work done by the battery to transfer charge $Q$ onto the 1st capacitor is $W=QV=\frac{Q^2}{C}$.</p>
<p>The final charges on the capacitors are $\frac12Q, \frac14Q, \frac{1}{16}Q, \frac{1}{64}Q, ...$ The total energy stored in all the capacitors at this time is $$E_{\infty}=\frac{Q^2}{2C}[(\frac12)^2+(\frac14)^2+(\frac{1}{16})^2+(\frac{1}{64})^2 +...]=\frac{Q^2}{6C}$$ The total loss of energy after closing S1 is $$W-E_{\infty}=\frac{5Q^2}{6C}=\frac56CV^2=\frac{20}{6}\pi\epsilon_0 RV^2$$</p>
<p>So I think option A should be the correct answer. The marking scheme appears to be wrong. </p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3460/net-energy-loss-in-infinite-series-of-capacitors&show=3469#a3469Thu, 16 May 2019 18:46:46 +0000Answered: Electric field for volume charge distributed slab
http://physics.qandaexchange.com/?qa=3455/electric-field-for-volume-charge-distributed-slab&show=3463#a3463
<p>Because the slab extends infinitely in the xy plane, the electric field lies only along the z direction. The differential form of Gauss' Law for such a one-dimensional electric field is $$\frac{dE}{dz}=\frac{\rho(z)}{\epsilon_0}$$ This can be integrated, using the boundary condition that the electric field must be zero at large distances from the slab, because it is electrically neutral.</p>
<hr>
<p>A more elementary way of solving the problem is to divide the slab into infinitesimally thin layers of thickness $dz$. The electric field of each layer is uniform, independent of distance from the layer, and is $dE=\frac{d\sigma}{2\epsilon}$ pointing away from the layer on each side for +ve surface charge density $d\sigma=\rho(z)dz$. The total electric field at any point inside or outside of the slab is found by superposition of fields from every such layer in the slab.</p>
<p><strong>Note</strong> that the total electric field at any point due to all layers which are closer to the centre plane $z=0$, is zero. This is because the charge density is anti-symmetric : for every layer of +ve area charge density on one side of $z=0$ there is a layer of -ve charge with the same magnitude of area density on the other side of $z=0$. The electric fields of these two layers cancel out for points which outside of the two layers, in the same way that the total electric field is zero outside of a parallel plate capacitor (if the plate dimensions are very much bigger than the distance from them). </p>
<p>From this observation you can see that the electric field outside of the slab is zero, because all layers in the slab are closer to the centre plane.</p>
<p>The simplest way of getting the field <strong>inside</strong> the slab is to apply Gauss' Law using a "pill box" Gaussian surface which has one face A of area S at the surface of the slab $z=a$ (where $E(a)=0$) and the other face B at distance $|z|<a$ from the centre plane. The other face(s) of the pill box are parallel to the z direction so the electric flux through them is zero. There is no electric flux through face A; the flux through face B is $E(z)S=\frac{q}{\epsilon_0}$ where by integration $$q=\int_a^z S\rho(z)dz$$ is the total charge inside the pill box. </p>
<p>See <a rel="nofollow" href="https://physics.stackexchange.com/q/196936">Electric field in a non-uniformly charged sheet</a>. An answer identical to mine is given in the duplicate question <a rel="nofollow" href="https://physics.stackexchange.com/q/197230">Finding the electric field of a NON uniform slab?</a></p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3455/electric-field-for-volume-charge-distributed-slab&show=3463#a3463Wed, 15 May 2019 21:58:01 +0000Find the equation of wave
http://physics.qandaexchange.com/?qa=3461/find-the-equation-of-wave
<blockquote><p>The linear mass density of a nonuniform wire under constant tension decreases gradually along the wire so that an incident wave is transmitted without reflection. The wire is uniform for $-\infty < x < 0$. In this region, a transverse wave has the form $$y(x, t) = 0.003 \cos (25x — 50t)$$ where $y$ and $x$ are in meters and $t$ is in seconds. In the region $0 < x < 20$ the linéar mass density decreases gradually from $\mu$ to $\frac14 \mu$. For $20 < x < +\infty$ the linear mass density is $\frac14 \mu$. Then prove that for $x > 20$ the wave equation is $$ y(x,t) = 0.0042 \cos (12.5 x — 50 t)$$</p>
</blockquote>
<p><img src="https://i.imgur.com/FkvLXz0.jpg" alt=""></p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3461/find-the-equation-of-waveWed, 15 May 2019 02:23:13 +0000Answered: To solve the equations of motion of a simple Pendulum $\ddot \theta = -\dfrac gl \sin \theta$
http://physics.qandaexchange.com/?qa=3454/solve-equations-motion-simple-pendulum-theta-dfrac-theta%24&show=3459#a3459
<p>You left out a minus sign in the 2nd equation. It should be written $$\ddot \theta \approx -\frac{g}{l}\theta$$ Now you can see that $\theta=e^{\sqrt{\frac{g}{l}} t}$ is not a solution because $\ddot \theta =+\frac{g}{l}\theta$. Instead we need a solution of the form $\theta=\pm e^{i\sqrt{\frac{g}{l}}t}$. Then we get $$\ddot \theta = i^2\frac{g}{l}e^{i\sqrt{\frac{g}{l}}t}=-\frac{g}{l}\theta$$ </p>
<p>Solutions to $$\ddot \theta = \pm\frac{g}{l}\theta$$ have the form $\theta=e^{\pm \sqrt{\frac{g}{l}} t}$ when the sign on the right is +ve and $\theta=e^{\pm i\sqrt{\frac{g}{l}} t}$ when the sign on the right is -ve. The most general solution is a linear combination : $$\theta_{+}=Ae^{+\sqrt{\frac{g}{l}} t}+ Be^{-\sqrt{\frac{g}{l}}t}=C\cosh(\sqrt{\frac{g}{l}} t+\phi)$$ $$\theta_{-}=Ae^{i\sqrt{\frac{g}{l}}t}+Be^{-i\sqrt{\frac{g}{l}} t}=C\cos(\sqrt{\frac{g}{l}} t+\phi)$$ for suitable values of $C, \phi$.</p>
<hr>
<p>Your 2nd equation $$\ddot \theta \approx -\frac{g}{l}\theta$$ is linear and is called the <strong>Harmonic Equation</strong>. Solutions are simply linear combinations of exponentials or trigonometric functions.</p>
<p>Your 1st equation $$\ddot \theta \approx -\frac{g}{l}\sin\theta$$ is non-linear. Solutions cannot be expressed in terms of simple exponential and trigonometric functions. They can be expressed in terms of <strong>elliptic functions</strong> as shown in <a rel="nofollow" href="https://www.math24.net/nonlinear-pendulum/">https://www.math24.net/nonlinear-pendulum/</a>.</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3454/solve-equations-motion-simple-pendulum-theta-dfrac-theta%24&show=3459#a3459Tue, 14 May 2019 01:48:52 +0000Answered: Thermal stress in a bar.
http://physics.qandaexchange.com/?qa=3450/thermal-stress-in-a-bar&show=3451#a3451
<p>Young's Modules is defined as,</p>
<p>$Y = \frac{\sigma}{\epsilon}$</p>
<p>where $\sigma$ is the strain defined as $\sigma = F/A$ and $\epsilon$ is the stress defined as $\epsilon = \frac{\Delta L}{L_0}$.</p>
<p>In this case we require the stress, thus re-arranging the first equation gives,</p>
<p>$\sigma = \epsilon Y$</p>
<p>After plugging in the definition of $\epsilon$ we then arrive at,</p>
<p>$\sigma = \frac{\Delta L}{L_0}Y$</p>
<p>Now as you already worked out, $\Delta L = L_0\alpha\Delta T$. Therefore after plugging this in, we arrive at the required result:</p>
<p>$\sigma = \frac{L_0\alpha\Delta T}{L_0}Y = \alpha\Delta T Y$</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3450/thermal-stress-in-a-bar&show=3451#a3451Tue, 30 Apr 2019 11:18:41 +0000Answered: Container filled with fluid move with a acceleration a
http://physics.qandaexchange.com/?qa=3446/container-filled-with-fluid-move-with-a-acceleration-a&show=3449#a3449
<p>Your calculation is correct so far but it is not complete.</p>
<p>In the vertical direction there is a gravitational field of strength $g$ acting downwards. This is equivalent to an acceleration upwards at rate $g$. The acceleration of the ball relative to the container is $a_y=3g$ upwards.</p>
<p>In the horizontal direction the ball accelerates to the right at $a_x$ <em>relative to the container</em>. This is different from the acceleration $a$ of the container itself, which is equivalent to a gravitational field acting to the left. So we can think of the container as stationary and acted upon by gravitational fields downwards and to the left.</p>
<p>Horizontally the ball must cover twice the distance as vertically but in the same time. Since $s \propto a$ and $a_y=3g$ then we must have $a_x=6g$.</p>
<p>The final step which you have missed is to relate $a_x$ (the acceleration of the ball relative to the container) to the acceleration $a$ of the container itself. As in the vertical direction we have $a_x=3a=6g$. Therefore $a=2g$. </p>
<p>The time for the ball to reach the edge of the container is $\sqrt{\frac{L}{3g}}$ as you calculated. So the options B, C are correct.</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3446/container-filled-with-fluid-move-with-a-acceleration-a&show=3449#a3449Fri, 19 Apr 2019 05:54:57 +0000time required to pass electricity through electroplating bath
http://physics.qandaexchange.com/?qa=3447/time-required-pass-electricity-through-electroplating-bath
<p>The time required to pass 3600 Coulomb of electricity through an electroplating bath using a current of 200 mA is <strong><strong>__</strong></strong>_ hours</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3447/time-required-pass-electricity-through-electroplating-bathWed, 17 Apr 2019 03:09:40 +0000Answered: Increasing resistance with temperature.
http://physics.qandaexchange.com/?qa=3432/increasing-resistance-with-temperature&show=3445#a3445
<p>The power delivered to the bulb is $P=V^2/R$. Assuming there are no heat losses, the power delivered to the bulb is related to the temperature increase of the filament by $P=c\frac{dT}{dt}$ where $c$ is the heat capacity of the bulb and $T$ is the temperature increase. Following your suggestion I shall model the variation of resistance with temperature as $R=R_0e^{\alpha T}$ where temperature $T=0$ at time $t=0$. Then $$\frac{dt}{dT}=\frac{cR_0}{V^2}e^{\alpha T}$$ which can be integrated to get $$t=\frac{cR_0}{\alpha V^2}(e^{\alpha T}-1)$$ For small values of $T$ such that $\alpha T \ll 1$ we get the approximation $$t\approx \frac{cR_0}{V^2}T$$</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3432/increasing-resistance-with-temperature&show=3445#a3445Mon, 15 Apr 2019 17:28:00 +0000Answered: Minimum force required to rotate a lamina when there is friction
http://physics.qandaexchange.com/?qa=3433/minimum-force-required-to-rotate-lamina-when-there-friction&show=3444#a3444
<p>This is quite difficult to solve but the result is fairly simple. Possibly there is an easier solution but I do not see it.</p>
<p>The friction force between the lamina and the rough ground is spread uniformly over the lamina. The force per unit area is $\sigma=\frac{\mu W}{A}$ where $W, A$ are the weight and surface area of the lamina and $\mu$ is the coefficient of friction.</p>
<p>Using polar co-ordinates with the fixed vertex as origin, the element of area is $dA=rdrd\theta$. The force on the element is $dF=\sigma dA$. The moment about the origin of the friction force on this element is $dM=rdF$. </p>
<p>If $p$ is the closest distance from the origin of the side of the lamina which does not pass through the origin, then the maximum distance from the origin at polar angle $\theta$ is $p\sec\theta$. The total moment of friction about the vertex is $$M=\sigma \int \int_0^{p\sec\theta} r^2drd\theta=\frac13 \sigma p^3\int \sec^3\theta d\theta$$ $$=\frac16 \sigma p^3 [\sec\theta \tan\theta+\ln|\sec\theta+\tan\theta|]$$ </p>
<p>See <a rel="nofollow" href="https://en.wikipedia.org/wiki/Integral_of_secant_cubed">Integral of Secant Cubed</a>. The limits of integration are $\theta_1, \theta_2$.<br>
<img src="https://i.imgur.com/8vyYA3N.png" alt=""><br>
Suppose the triangle has vertices ABC with side lengths $a< b< c$. With the axis at vertex A we have $$p=b, \theta_1=0, \sec\theta_1=1, \tan\theta_1=0, \sec\theta_2=\frac{c}{b}, \tan\theta_2=\frac{a}{b}$$ $$M_A=\frac16 \sigma [abc+b^3\ln(\frac{a+c}{b})]$$ Likewise when vertex B is the axis $$M_B=\frac16 \sigma [abc+a^3\ln(\frac{b+c}{a})]$$ For vertex C $$p=ab/c, \sec\theta_1=a/p, \tan\theta_1=a^2/cp, \sec\theta_2=b/p, \tan\theta_2=b^2/cp$$ $$c^3M_C=\frac16 \sigma [b^3(abc+a^3\ln(\frac{b+c}{a})+a^3(abc+b^3\ln(\frac{a+c}{b}))]=a^3M_A+b^3M_B$$ </p>
<p>The moment of friction is balanced by the moment of the applied force. When the axis is at vertices A, B the applied force is minimum when this force is applied at B, A respectively and it is perpendicular to AB; the moments are then $M_A=cF_A, M_B=cF_B$. When the axis is at vertex C the applied force is minimum when applied at A perpendicular to AC; then $M_C=bF_C$. Therefore $$bc^2F_C=a^3F_A+b^3F_B$$</p>
<p>In the question $\theta$ equals angle A so $$F_C=F_A\tan\theta \sin^2\theta +F_B\cos^2\theta $$</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3433/minimum-force-required-to-rotate-lamina-when-there-friction&show=3444#a3444Fri, 12 Apr 2019 21:02:01 +0000Answered: Calculating equivalent capacitance.
http://physics.qandaexchange.com/?qa=3442/calculating-equivalent-capacitance&show=3443#a3443
<p>The middle diagram shows the correct equivalent configuration. </p>
<p>The diagram on the right assumes that the potential is the same at all points of the interface between the upper and lower dielectrics. If this is true then the interface can indeed be replaced by a conducting plate and the upper dielectric becomes a single capacitor, as suggested in the right-hand diagram. </p>
<p>However, the interface is a dielectric surface not a conducting surface, so it need not be at the same potential at all points adjacent to each of the two lower dielectrics. The lower surface of the upper dielectric can be split into two surfaces which can be at different potentials. This is recognised in the middle diagram. </p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3442/calculating-equivalent-capacitance&show=3443#a3443Fri, 12 Apr 2019 18:12:45 +0000Answered: Velocity of image of submerged object when liquid surface moves
http://physics.qandaexchange.com/?qa=3437/velocity-image-submerged-object-when-liquid-surface-moves&show=3440#a3440
<p>Suppose the observer is a fixed distance $H$ above the coin. Then the distance from the observer to the water surface is $H-h$ and the apparent distance of the coin below the surface is $\frac34 h$. So the apparent distance of the coin from the observer (measuring downwards) is $$y=(H-h)+\frac34h=H-\frac14h$$</p>
<p>The velocity of the water surface is $\frac{dh}{dt}=+8m/s$ (plus because $h$ is increasing) so the rate at which the image of the coin is moving downwards (in the direction of increasing y) is $$\frac{dy}{dt}=-\frac14\frac{dh}{dt}=-2m/s$$ The minus sign indicates that the image is moving upwards towards the observer, whereas y increases downwards.</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3437/velocity-image-submerged-object-when-liquid-surface-moves&show=3440#a3440Mon, 08 Apr 2019 16:12:29 +0000Given the vector current density, determine the total current flowing outward through a circular band
http://physics.qandaexchange.com/?qa=3431/current-density-determine-current-flowing-outward-circular
<p>Given the vector current density, determine the total current flowing outward through a circular band.:</p>
<p><a rel="nofollow" href="https://i.stack.imgur.com/klQAo.png"><img src="https://i.stack.imgur.com/klQAo.png" alt="enter image description here"></a></p>
<p>The answer should be 518A. It comes something around 3255 A. Where is the mistake?</p>
<p><a rel="nofollow" href="https://i.stack.imgur.com/g1T1l.jpg"><img src="https://i.stack.imgur.com/g1T1l.jpg" alt="enter image description here"></a></p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3431/current-density-determine-current-flowing-outward-circularFri, 29 Mar 2019 11:20:51 +0000Nucleon-Nucleon Scattering
http://physics.qandaexchange.com/?qa=3430/nucleon-nucleon-scattering
<p><img src="https://i.imgur.com/hZhT97F.png[/img]" alt=""><br>
<img src="https://i.imgur.com/rklrj0w.jpg[/img]" alt=""></p>
<p>I computed the orbital angular momentum classically, and then used quantum mechanics. But I am not convinced of what I got, because $L \alpha \sqrt{E}$ is a proportional equation that does not justify that when E < 20 MeV you get L ~ 0 (I do not see, numerically speaking, the difference between either plugging 20 MeV or 19MeV).</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3430/nucleon-nucleon-scatteringWed, 27 Mar 2019 12:38:33 +0000Answered: How to apply the gauss law when charge density is a function of not only $r$?
http://physics.qandaexchange.com/?qa=3416/how-apply-the-gauss-law-when-charge-density-function-not-only&show=3428#a3428
<p>In general the electric potential (or electric field) at point P outside of a charge distribution can be expanded as a power series known as a <strong>multipole expansion</strong> : $$V(z)=\frac{A}{r}+\frac{B}{r^2}+\frac{C}{r^3}+\frac{D}{r^4}+...$$ where the coefficients $A, B, C, D, ...$ depend on the spherical polar and azimuthal co-ordinates $\theta, \phi$ and $r$ is the distance of P from centre O of the charge distribution. See <a rel="nofollow" href="http://physics.qandaexchange.com/?qa=3393/electric-field-far-away-of-a-bunch-of-charges">Electric field far away from a bunch of charges</a>.</p>
<p>In this case the total charge on the sphere is zero : $$Q=\int_0^R \int_0^{\pi} \rho r^2\sin\theta d\theta dr=KR\int_0^R (R-2r) dr \int_0^{\pi}\sin^2\theta d\theta=0$$ because the integral wrt $r$ is zero. So the electric potential far from the centre of the sphere will not be proportional to $1/r$ as it is for a point charge : that is, $A=0$ here. </p>
<p>It is convenient to divide the sphere into rings such as A and B which are parallel to the xy plane, because such rings have uniform charge density. A typical ring has radius $y=r\sin\theta$ and each point on it is the same distance $s$ from the point of interest P which lies on the $z$ axis. The potential at P due to the ring is simply $V=kQ/s$ where $k$ is Coulomb's constant,</p>
<p><img src="https://i.imgur.com/rIu95cR.png" alt=""></p>
<p>Using the Cosine Rule for ring A $$s^2=z^2-2rz\cos\theta+r^2=z^2(1-x)$$ where $x=p(c-p), c=2\cos\theta, p=\frac{r}{z}$. The charge on each ring is $$dQ=2\pi y \rho r d\theta dr=2\pi\rho r^2\sin\theta d\theta dr$$ </p>
<p>Ring B has the same charge as ring A and is located symmetrically about O, the centre of the sphere. The potential at P due to both rings A and B, each with charge $dQ$, is $$dV=\frac{dQ}{4\pi \epsilon_0 }(\frac{1}{s_1}+\frac{1}{s_2})=\frac{KR(R-2r)}{2\epsilon_0 z}(\frac{1}{\sqrt{1- x}}+\frac{1}{\sqrt{1+ x}})\sin^2\theta d\theta dr$$ </p>
<p>The reciprocals can be expanded as power series : $$\frac{1}{\sqrt{1-x}}=1+\frac12x+\frac38x^2+\frac{5}{16}x^3+\frac{35}{128}x^4+\frac{63}{256}x^5+\frac{231}{1024}x^6+...$$ $$\frac{1}{\sqrt{1+x}}=1-\frac12x+\frac38x^2-\frac{5}{16}x^3+\frac{35}{128}x^4-\frac{63}{256}x^5+\frac{231}{1024}x^6+...$$ which have the half-sum $$\frac12 (\frac{1}{\sqrt{1-x}}+\frac{1}{\sqrt{1+x}})=1+\frac38x^2+\frac{35}{128}x^4++\frac{231}{1024}x^6+...$$ </p>
<p>The advantage of taking 2 rings together is that we have halved the number of terms in the series. This simplifies calculation. To avoid double-counting we must reduce the range of integration to $\theta=0 \to \pi/2$.</p>
<p>The words "far from the centre" suggest that we should assume $z\gg R$ and therefore we ignore all except the lowest power terms. Substituting from above $$x^2=p^2(c-p)^2\approx \frac{r^2}{z^2}\cos^2\theta$$ </p>
<p>The potential at P is approximately $$V\approx \frac{KR}{\epsilon_0 z} \int_0^{\pi/2} d\theta \int_0^R dr . (R-2r)(1+\frac38 \frac{r^2}{z^2} \cos^2 \theta ) \sin^2\theta$$ $$= \frac{KR}{\epsilon_0 z} \int_0^R (R-2r)\frac38 \frac{r^2}{z^2} dr \int_0^{\pi/2} \cos^2 \theta \sin^2\theta d\theta$$ $$= \frac{KR}{\epsilon_0 z} (-\frac{R^4}{16z^2} )(\frac{\pi}{16})$$ $$= -\frac{\pi KR^5}{256\epsilon_0 z^3}$$ The electric field at point P is $$E_z=-\frac{dV}{dz}=-\frac{3\pi KR^5}{256\epsilon_0 z^4}$$</p>
<p><strong>Notes :</strong> </p>
<ol>
<li><p>We should expect the potential to be -ve because the outer part of the sphere (which is closest to P) is negatively charged. </p>
</li>
<li><p>The dominant term in the charge distribution is not a <em>dipole</em> $V\propto 1/r^2$ but a <em>quadrupole</em> $V\propto 1/r^3$. We should expect to lose the dipole term because the centres of +ve and -ve charge coincide. </p>
</li>
<li><p>This result gives only the leading term in the <strong>multipole expansion</strong> and applies only for $z\gg R$. We could obtain more terms by retaining powers of $x^4, x^6...$ in the half-sum expansion and higher powers of $r$ when substituting for $x$. As more terms are retained the result becomes more accurate for smaller values of $z \gt R$.</p>
</li>
</ol>
Physics Problemshttp://physics.qandaexchange.com/?qa=3416/how-apply-the-gauss-law-when-charge-density-function-not-only&show=3428#a3428Tue, 26 Mar 2019 16:51:47 +0000Where do I make the approximations in this proton and electron collision problem?
http://physics.qandaexchange.com/?qa=3407/where-make-approximations-proton-electron-collision-problem
<blockquote><p>A proton moving with a velocity $\beta c$ collides with a stationary electron of mass $m$ and knocks it off at an angle $\theta$ with the incident direction. Show that the energy imparted to the electron is approximately $$T \approx \frac{2\beta^2 \cos^2\theta}{1-\beta^2\cos^2\theta}mc^2$$</p>
</blockquote>
<p><em>Link to question :</em><br>
<a rel="nofollow" href="https://www.chegg.com/homework-help/questions-and-answers/proton-moving-velocity-collides-stationary-electron-mass-m-knocks-angle-incident-direction-q35643739?trackid=VRP2b2Jt">https://www.chegg.com/homework-help/questions-and-answers/proton-moving-velocity-collides-stationary-electron-mass-m-knocks-angle-incident-direction-q35643739?trackid=VRP2b2Jt</a></p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3407/where-make-approximations-proton-electron-collision-problemSun, 17 Mar 2019 04:34:00 +0000Answered: Maximum charge and maximum current
http://physics.qandaexchange.com/?qa=3406/maximum-charge-and-maximum-current&show=3414#a3414
<p>This problem can be solved by two methods : (A) using conservation of flux linkage and energy, and (B) using differential equations.</p>
<p>What you are missing in your solution is the conservation of energy. No energy is lost because there is no resistance in the circuit after the battery is disconnected. </p>
<p>(<strong>Note :</strong> With the type of switch depicted in the diagram, the circuit is broken - although only for a short time. The current through $L_1$ cannot then flow. This causes a large back emf and a spark across the switch, losing most of the energy stored in the inductor. To avoid loss of energy the switch needs to be smooth, with a small amount of overlap, so that there is never a break in the circuit containing the inductor.)</p>
<p><strong>How the Circuit Behaves</strong></p>
<p>While the switch is in position 1 there is a current $i_0=E/r$ flowing through $L_1$ and (presumably) none through $L_2$. It is possible that a sinusoidal current flows around circuit QRUT before the switch is thrown. However, we are not told anything about this so we must assume that there was initially no current in $L_2$.</p>
<p>When the switch is changed to position 2 the current $i_0$ continues to flow through $L_1$ - it cannot change instantaneously. For the same reason current cannot flow through $L_2$ immediately - it takes time to build up. Instead, all of the current from $L_1$ flows initially onto the capacitor. </p>
<p><img src="https://i.imgur.com/ZZgJrs6.png" alt=""></p>
<p>There is initially no voltage across either inductor or the capacitor. As charge $q$ builds up on the capacitor there is an increasing voltage across all 3 components. This voltage increases the (upward) current $i_2$ through $L_2$ and decreases the (downward) current $i_1$ through $L_1$. </p>
<p>When the charge (and voltage) on the capacitor reaches a maximum, no more current flows into the capacitor. At this instant the current through $L_2$ is the same as that through $L_1$ - ie $i_1=i_2$ - because no current flows into or out of circuit PQTS. You can find the (maximum) charge on the capacitor at this instant using conservation of energy.</p>
<p>Although the charge and voltage on the capacitor then both decrease, this voltage continues to increase the (upward) current in $L_2$ and to decrease the (downward) current in $L_1$ until the charge on the capacitor becomes zero again. The current through $L_2$ is then a maximum because the polarity of the capacitor then switches, the voltage on it reverses, and this reduces the (upward) current $i_2$ while also increasing downward current $i_1$. By applying the conservation of energy again you can find the maximum current through $L_2$.</p>
<p><strong>(A) Solution using Conservation of Energy</strong></p>
<p>When the switch is thrown the current through $L_1$ is $i_0=E/r$. (This notation differs from yours.) Flux linkage is conserved in loop PQTS because this loop contains only inductors. When the currents in $L_1, L_2$ are generally $i_1, i_2$ we can write $$L_1i_0=L_1i_1+L_2i_2$$ </p>
<p>When the charge on the capacitor is a maximum $q_0$ then $i_1=i_2$ so $(L_1+L_2)i_1=L_1i_0$. By conservation of energy we can write $$L_1i_0^2=L_1i_1^2+L_2i_2^2+\frac{1}{C}q^2=(L_1+L_2)i_1^2+\frac{1}{C}q_0^2=\frac{L_1^2}{L_1+L_2}i_0^2+\frac{1}{C}q_0^2$$ $$q_0^2=\frac{CL_1L_2}{L_1+L_2}i_0^2$$ As we shall find later we can write this in the form $$q_0=\frac{i_0}{\omega}$$ where $\omega$ is the angular frequency of oscillations in the circuit.</p>
<p>When the charge on the capacitor becomes zero again ($q=0$) then the current through $L_2$ reaches its maximum value $i_3$. Equations for the conservation of flux linkage and energy then become $$L_1i_0=L_1i_1+L_2i_3$$ $$L_1i_0^2=L_1i_1^2+L_2i_3^2=L_1i_0^2-2L_2i_0i_3+\frac{L_2^2}{L_1}i_3^2+L_2i_3^2$$ $$0=[(\frac{L_2}{L_1}+1)i_3-2i_0]L_2i_3$$ We have already met one solution $i_3=0$; the other solution is $$i_3=\frac{2L_1}{L_1+L_2}i_0$$ We can go on to find the <strong>minimum current</strong> through $L_1$ which is $$i_4=\frac{L_1-L_2}{L_1+L_2}i_0$$</p>
<p><strong>(B) Solution using Differential Equations</strong></p>
<p>The voltage across the components in the circuit are related to the currents and charges by $$V=-L_1\frac{di_1}{dt}=L_2\frac{di_2}{dt}=\frac{1}{C}q$$ $$i_1-i_2=\frac{dq}{dt}$$ From these we get the equations $$L_1i_1+L_2i_2=L_1i_0$$ $$\frac{d^2q}{dt^2}+\omega^2 q=0$$ $$\omega^2=\frac{1}{C}(\frac{1}{L_1}+\frac{1}{L_2})$$ with the solutions $$q=q_0\sin\omega t$$ $$(L_1+L_2)i_1=L_1i_0+L_2\frac{dq}{dt}=L_1i_0+L_2\omega q_0\cos\omega t$$ $$(L_1+L_2)i_2=L_1i_0-L_2\frac{dq}{dt}=L_1i_0-L_1\omega q_0\cos\omega t$$ Substitute $i_1=i_0$ when $t=0$ into the 1st eqn to get $$i_0=\omega q_0$$ We can then write general solutions for the currents as $$i_1=i_0\frac{L_1}{L_1+L_2}(1+\frac{L_2}{L_1}\cos\omega t)$$ $$i_2=i_0\frac{L_1}{L_1+L_2}(1-\cos\omega t)$$ The minimum current through $L_1$ and the maximum current through $L_2$ are respectively $$i_4=\frac{L_1-L_2}{L_1+L_2}i_0$$ $$i_3=\frac{2L_1}{L_1+L_2}i_0$$</p>
<p><img src="https://i.imgur.com/OBmwT1g.png" alt=""></p>
<p><strong>Notes :</strong></p>
<ol>
<li><p>The current $i_2$ through $L_2$ always has a minimum of zero and is always positive (upward in the diagram). </p>
</li>
<li><p>Unlike $i_2$, the current $i_1$ through $L_1$ <strong>can become negative</strong> (flow upwards in the diagram). This happens if $L_1 \lt L_2$.</p>
</li>
<li><p>The charge on the capacitor is a maximum (both polarities) when the same current flows through both inductors - ie $i_1=i_2$.</p>
</li>
<li><p>When this happens the current through both inductors equals half the maximum current through $L_2$ - ie $$i_1=i_2=\frac{L_1}{L_1+L_2}i_0=\frac12 i_3$$ </p>
</li>
</ol>
Physics Problemshttp://physics.qandaexchange.com/?qa=3406/maximum-charge-and-maximum-current&show=3414#a3414Sat, 16 Mar 2019 18:41:11 +0000Answered: Problem 5.13 Griffiths; Balancing magnetic attraction with electrical repulsion
http://physics.qandaexchange.com/?qa=3412/griffiths-balancing-magnetic-attraction-electrical-repulsion&show=3413#a3413
<p>The line charges must have some mass. Because of this it is impossible to accelerate them to the speed of light. </p>
<p>Increasing their speed also increases their inertia. Initially almost all of the work done increases their speed and almost none increases their inertia. As their speed approaches the speed of light, almost all of the work done on them increases their inertia and there is little increase in speed.</p>
<p>So their speed will always be less than the speed of light, although you can get as close to $c$ as you wish.</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3412/griffiths-balancing-magnetic-attraction-electrical-repulsion&show=3413#a3413Thu, 14 Mar 2019 19:29:11 +0000Griffiths 5.4 Force on a square loop
http://physics.qandaexchange.com/?qa=3404/griffiths-5-4-force-on-a-square-loop
<p><img src="https://i.imgur.com/HckbPJf.png[/img]" alt=""></p>
<p><img src="https://i.imgur.com/9SEg3Li.png[/img]" alt=""></p>
<p>Before doing any calculation, I see the net force being zero;</p>
<p>$$ F_{AB} = -F_{CD}$$</p>
<p>$$ F_{BC} = -F_{DA}$$ </p>
<p>But apparently this is not the case. The provided solution is:</p>
<p><img src="https://i.imgur.com/75ijztI.png[/img]" alt=""></p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3404/griffiths-5-4-force-on-a-square-loopSat, 09 Mar 2019 11:53:06 +0000How to induce large current in flexible wire and explaining how does it stretch into a circle
http://physics.qandaexchange.com/?qa=3401/induce-large-current-flexible-explaining-does-stretch-circle
<blockquote><p>The idea is to explain how could we induce current in a flexible loop using an initially free-of-current wire.</p>
</blockquote>
<p><strong>Some background.</strong></p>
<p>The current has to be the same all the way around (otherwise charge would be gathered at some point and in that accumulation of charge, the electric field would point in a way that the flow would even out). </p>
<p>The two forces involved in driving the current around the loop are the source force (typically a battery) and the electrostatic force, which smooths out the flow. The line integral of the vector sum of the previous forces is known as electromotive force (emf).</p>
<p><strong>My answer to the problem</strong> </p>
<p>My difficulty in this problem is that I do not have clear the sketch. I would say that the following happens:</p>
<p>Imagine that the loop is in the xy plane. The velocity of the current is tangential to the loop and its direction is clockwise; $B$ will point in the $-z$ direction and the force will point radially outwards. This force would be the source one (provided by the battery and the one that does work to move the loop; I know magnetic force does no work; the individual radial vectors of the force do work but the net work done by all of them is zero) if the wire were just connected to a battery. I also understand that the net electrostatic force is zero (this is a closed path).</p>
<p>But here there is no battery, the current is induced by a change in the current of the wire</p>
<p>My intuition tells me that is the change in the current of the wire what changes the flux of the wire and the changing flux induces the emf in the loop. But not really sure of this...</p>
<p><strong>What's the nature of the induced current in the loop?</strong></p>
<p><strong>EDIT:</strong> The original exercise does not provide details on the set up, but my interpretation was the following: we start with two free-of-current wires: one is a loop, the other a straight line. We apply current on the straight wire, which induces an emf on the loop (which will stretch into a circle; this idea came from one of the sources I checked: <a rel="nofollow" href="https://physics.stackexchange.com/questions/239591/magnetic-potential-energy">https://physics.stackexchange.com/questions/239591/magnetic-potential-energy</a> ).</p>
<p>Souces:</p>
<p><a rel="nofollow" href="https://physics.stackexchange.com/questions/239591/magnetic-potential-energy">https://physics.stackexchange.com/questions/239591/magnetic-potential-energy</a></p>
<p><a rel="nofollow" href="https://physics.stackexchange.com/questions/266895/what-force-causes-the-induced-emf-of-a-loop-and-the-difference-between-a-loop?rq=1">https://physics.stackexchange.com/questions/266895/what-force-causes-the-induced-emf-of-a-loop-and-the-difference-between-a-loop?rq=1</a></p>
<p>Introduction to Electrodynamics by David Griffiths</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3401/induce-large-current-flexible-explaining-does-stretch-circleFri, 08 Mar 2019 16:08:33 +0000Answered: Questions from a jumping kangaroo
http://physics.qandaexchange.com/?qa=3263/questions-from-a-jumping-kangaroo&show=3400#a3400
<p>A general algorithm which works for all functions $f(x)$ will have to use a <strong>numerical method</strong> of solution - ie start with an initial approximation to the trajectory, estimate how close this comes to the optimal solution (the "error"), use this to make an adjustment to the solution, estimate the new "error," and continue the same cycle until the "error" is small enough for your purposes.</p>
<p><strong>Analytic solutions</strong> in the form of closed functions are possible only for particular shapes of $f(x)$ such as rectangles, circles, and ellipses.</p>
<p>If the object is convex and has a vertical plane of symmetry then you can use the method in my answer to <a rel="nofollow" href="https://physics.stackexchange.com/questions/288193/">Minimum speed required to clear rectangular object</a>. This involves finding points A, B on the same horizontal level, at which the projectile could graze the obstacle at an angle of $45^{\circ}$ to the horizontal. The distance AB enables you to find the required speed at these points, and the height of AB above the ground enables you to find the minimum speed with which the object should be launched.</p>
<p>This appears to be the method which you are using in your calculation.</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3263/questions-from-a-jumping-kangaroo&show=3400#a3400Mon, 04 Mar 2019 16:03:59 +0000Answered: Force that one half of uniformly charged solid sphere exerts on other half - Griffith 2.43
http://physics.qandaexchange.com/?qa=3368/force-uniformly-charged-solid-sphere-exerts-other-griffith&show=3399#a3399
<p><img src="https://imgur.com/yKhoEy9" alt=""></p>
<blockquote><p>The problem asks to calculate the force on one part due to other. But the electric field I'm using to do this calculation is the result E of complete solid non-conducting sphere. <strong>So why are we considering E of the northern hemisphere to calculate force it experience from the southern hemisphere?</strong></p>
</blockquote>
<p>This is a very good question.</p>
<p>One way of finding the total electrostatic force on the N hemisphere is to calculate the vector force $F_{ij}=k q_i q_j / r_{ij}^2$ on every charge $q_i$ in the N hemisphere due to every charge $q_j$ in the S hemisphere, then take the vector sum of all forces $F_{ij}$ : </p>
<p>$$F_N=\sum_{i \in N} \sum_{j \in S} F_{ij}$$</p>
<p>This double sum or integral is very difficult to calculate, for two reasons : </p>
<ol>
<li><p>the general expression for the force $F_{ij}$ between every pair of 2 charges is a complicated function using the Cartesian co-ordinate system. even more so when using spherical co-ordinates ;</p>
</li>
<li><p>each of the 2 charges $q_i, q_j$ in every pair has 3 co-ordinates, so in total there will be a sum ranging over 6 possible co-ordinates. </p>
</li>
</ol>
<p>An alternative method of solution (used in Jorge Daniel's answer) is to sum the <strong>total</strong> electrical force $F_i$ on each charge $q_i$ in the N hemisphere due to <strong>all</strong> of the charges in <strong>both</strong> the S and N hemispheres : </p>
<p>$$F_N=\sum_{i \in N} F_i=\sum_{i \in N} \sum_{j \in N,S} F_{ij}$$ </p>
<p>Whereas $F_{ij}$ is a very complicated expression, the total force $F_i$ on each charge is very much simpler because it is known to be radial and depends only on the radial distance of the charge $q_i$ from the centre of the sphere. </p>
<p>As you rightly point out, this means that within the expression for $F_i$ we will include the force $F_{ij}$ on charge $q_i$ due to other charges $q_j$ which are <strong>also in the N hemisphere</strong>. </p>
<p>However, if charge $q_j$ is also within the N hemisphere ($j \in N$), then when we take the sum over all charges $q_i$ in the N hemisphere ($i \in N$), this sum will include not only $F_{ij}$ but also the equal and opposite force $F_{ji}$ acting on $q_j$ due to charge $q_i$. These 2 contributions to the total force on all particles in the N hemisphere will cancel out because $F_{ij}=-F_{ji}$. </p>
<p>On the other hand, if charge $q_j$ is in the S hemisphere ($j \in S$) then the force $F_{ji}$ will <strong>not</strong> be included in the first summation over $i \in N$.</p>
<p>See <a rel="nofollow" href="https://physics.stackexchange.com/questions/23071/">Find the net force the southern hemisphere of a uniformly charged sphere exerts</a>.</p>
<hr>
<p>To make the above explanation clearer, suppose we have a system of only 4 charges $q_i$ with $i=1 \to 4$. Between every pair of charges there is an electrical force $F_{ij}$ meaning the force that charge $j$ exerts on charge $i$. </p>
<p>Suppose that charges $i=1 , 2$ constitute one object (which we call "the N hemisphere") and charges $i=3, 4$ constitute a second object ("the S hemisphere"). Then the total force on the N hemisphere due to the S hemisphere is $$F_N=F_{13}+F_{14}+F_{23}+F_{24}$$ The total forces on charges $q_1, q_2$ due to <strong>all</strong> other charges in <strong>both</strong> hemispheres (ie summing over $j \in N, S$) are $$F_1=F_{12}+F_{13}+F_{14}$$ $$F_2=F_{21}+F_{23}+F_{24}$$ If we add the forces on charges $q_1$ and $q_2$ (summing over $i \in N$) we get $$F_N'=F_1+F_2=(F_{12}+F_{21})+F_{13}+F_{14}+F_{23}+F_{24}=F_N$$ because $F_{12}=-F_{21}$. </p>
<p>Thus both methods give the same answer.</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3368/force-uniformly-charged-solid-sphere-exerts-other-griffith&show=3399#a3399Mon, 04 Mar 2019 15:02:39 +0000question about 4-velocity
http://physics.qandaexchange.com/?qa=3396/question-about-4-velocity
<p>A particle is moving in the x, y plane at a speed of v = 0.80 and it is travelling an angle of 60 degrees<br>
above the x-axis.</p>
<p>(a) Rotate the spatial coordinates so that v lies along the x-axis and then construct the components of the 4-velocity for the particle.</p>
<p>(b) Construct the remaining orthonormal basis vectors for the particle in the rotated frame, then rotate the spatial coordinates back to find the basis vectors in the original frame.</p>
<p>(c) Perform the same steps above but this time begin by rotating the spatial coordinates so that <br>
v lies along the y-axis. Do you get the same result for the basis vectors after rotating back to the original frame? Should you? Draw a picture with ~v, the original (x, y) frame, and the two rotated frames to explain what is happening.</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3396/question-about-4-velocityWed, 27 Feb 2019 22:32:29 +0000Answered: Rotating ball inside rotating cylinder.
http://physics.qandaexchange.com/?qa=3359/rotating-ball-inside-rotating-cylinder&show=3395#a3395
<p>I also do not understand the textbook solution.</p>
<p>In an equilibrium situation the ball is effectively rolling down a plane inclined at angle $\theta$ while the plane is itself accelerating upwards along the plane. In the ground frame of reference the centre of mass of the ball is stationary, so the resultant force on it must be zero. However, there is a non-zero torque which causes rotational acceleration down the plane.</p>
<p>The torque acting on the ball is $mgr\sin\theta$ where $r$ is its radius. Its acceleration down the plane, relative to the point of contact, is $r\beta$ where $\beta$ is its angular acceleration. Since there is no slipping at the point of contact, $r\beta$ must also equal the acceleration of the point of contact up the plane, which is $R\alpha$ where $R>r$ is the radius of the cylinder. That is : $$r\beta=R\alpha$$ </p>
<p>The equation of motion for the angular acceleration of the ball is therefore $$mgr\sin\theta=I\beta=\frac25 mr^2 \frac{R}{r}\alpha$$ $$\sin\theta=\frac{2R\alpha}{5g}$$</p>
<p>If there is a finite coefficient of friction $\mu$ then we must also have that $$\tan\theta \le \mu$$</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3359/rotating-ball-inside-rotating-cylinder&show=3395#a3395Wed, 27 Feb 2019 00:00:45 +0000Answered: Electric field far away from a bunch of charges
http://physics.qandaexchange.com/?qa=3393/electric-field-far-away-from-a-bunch-of-charges&show=3394#a3394
<p>You have not defined the problem well. "A bunch of charges" is a vague description. Both answers could be correct, or neither, depending on the arrangement of charges.</p>
<p>The electric field outside of any general system of charges can be modelled as a <a rel="nofollow" href="http://farside.ph.utexas.edu/teaching/jk1/lectures/node32.html"><strong>multi-pole expansion</strong> </a> of the form $$E=k(\frac{q_0}{r^2}+\frac{q_1}{r^3}+\frac{q_2}{r^4}+...)$$ The coefficients $q_k$ depend on the polar and azimuthal angles $\theta, \phi$ and are related to the <strong>spherical harmonic functions</strong>. The 1st term represents a monopole with the resultant charge on the bunch. The 2nd and 3rd terms represent a dipole and quadrupole respectively.<br>
<img src="https://i.imgur.com/dOpRZgf.png" alt=""><br>
If the "bunch of charges" has an overall non-zero charge $q_0$ then far from the bunch the 1st term will dominate. It will resemble a point charge with an electric field which varies approximately as $1/r^2$.</p>
<p>However, if the bunch has an overall zero charge $q_0=0$ but the centres of +ve and -ve charge do not coincide then the 2nd term will dominate. The electric field far from the bunch is will resemble that of a dipole, which varies as $1/r^3$.</p>
<p>If the centres of +ve and -ve charge do coincide, then the 2nd term will also vanish ($q_1=0$) - for example, 4 charges at the corners of a square, alternating. This arrangement could have a non-zero 3rd term (a quadrupole, $q_2 \ne 0$), for which the electric field is proportional to $1/r^4$.</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3393/electric-field-far-away-from-a-bunch-of-charges&show=3394#a3394Mon, 25 Feb 2019 02:07:43 +0000Answered: Electric field intensity at the centre of a solid hemisphere with uniform volume charge density
http://physics.qandaexchange.com/?qa=3360/electric-intensity-centre-hemisphere-uniform-volume-density&show=3392#a3392
<p>The question is contradictory : It says that the charge density is uniform (the same at every point in the hemisphere), but also that it depends on azimuthal co-ordinate $\phi$, with $\rho=3\phi$. The latter is a very unusual variation, because it has a discontinuity at the meridian $\phi=0, 2\pi$. Perhaps a polar variation is intended, with $\rho=3\theta$, which has no discontinuity.</p>
<p><strong>(a) Uniform Volume Charge Density $\rho=\text{const}$</strong></p>
<p>Divide the hemisphere into rings (hoops) of thickness $dr$ and width $rd\theta$ located at constant values of $r, \theta$ parallel to the xy plane, with the centre as origin. The radius of the ring is $y=r\sin\theta$, its volume is $dV=2\pi y r d\theta dr$ and its charge is $dq=\rho dV$. </p>
<p>The z-component of the electric field due to this ring at the origin is $dE_z=-kdq \cos\theta/r^2$. The total electric field due to all rings in the hemisphere is $$E_z=-\int dE_z=-k\pi \rho \int_0^a dr \int_0^{\pi/2} \sin2\theta d\theta =-\frac{\rho a}{4\epsilon_0} $$ <br>
The total charge on the hemisphere is $Q=\frac23 \pi \rho a^3$ so the electric field can be re-written as $$E_z=-\frac{3Q}{8\pi \epsilon_0 a^2}$$</p>
<p><strong>(b) Polar Charge Density $\rho=3\theta$</strong></p>
<p>Again we can use rings because they have a constant value of $\theta$.</p>
<p>$$E_z=-\int 3\theta dV=-k\pi \int_0^a dr \int_0^{\pi/2} 3\theta \sin2\theta d\theta =-\frac{a}{4\epsilon_0} \frac34 [\sin2\theta-2\theta\cos2\theta]_0^{\pi/2} =-\frac{3\pi a}{16\epsilon_0}$$ </p>
<p>The total charge on the hemisphere is $$Q=\int dq=\pi\int_0^a 3r^2 dr \int_0^{\pi/2} \theta\sin2\theta.d\theta=\frac14 \pi^2 a^3$$ so the electric field can be re-written as $$E_z=-\frac{3Q}{4\pi \epsilon_0 a^2}$$</p>
<p><strong>(c) Azimuthal Charge Density $\rho=3\phi$</strong></p>
<p>We can no longer use rings because $\phi$ is not constant for a ring. The volume element is $dV=r^2\sin\theta dr d\theta d\phi$ and the charge on each element is $dq=3\phi dV$. </p>
<p>The electric field at the centre due to an element of charge at ($r,\theta,\phi$) has components $$dE_z=-k\frac{dq}{r^2}\cos\theta, dE_y=-k\frac{dq}{r^2}\sin\theta\cos\phi, E_x=-k\frac{dq}{r^2}\sin\theta\sin\phi$$ so we get $$E_z=-3k\int_0^a dr \int_0^{\pi/2} \frac12 \sin2\theta d\theta \int_0^{2\pi} \phi d\phi=-\frac{3\pi a}{2\epsilon_0}$$ $$E_y=-3k\int_0^a dr \int_0^{\pi/2} \frac12 (1-\cos2\theta) d\theta \int_0^{2\pi} \phi \cos\phi d\phi=-\frac{3 a}{2\epsilon_0}(\frac{\pi}{4}+1)$$ $$E_x=-3k\int_0^a dr \int_0^{\pi/2} \frac12 (1-\cos2\theta) d\theta \int_0^{2\pi} \phi \sin\phi d\phi=-\frac{3a}{2\epsilon_0}(\frac{\pi}{4}+1)$$<br>
The total charge on the hemisphere is $$Q=\int dq=\int_0^a 3r^2 dr \int_0^{\pi/2} \sin\theta d\theta \int_0^{2\pi} \phi d\phi=2\pi^2 a^3$$ so we can express the components of the field as $$E_z=-\frac{3Q}{4\pi \epsilon_0 a^2}, E_y=E_x=-\frac{3Q}{4\pi \epsilon_0 a^2}(\frac14+\frac{1}{\pi})$$</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3360/electric-intensity-centre-hemisphere-uniform-volume-density&show=3392#a3392Sat, 23 Feb 2019 23:06:15 +0000Answered: What is the fraction of work per second by F is converted into heat.
http://physics.qandaexchange.com/?qa=3373/what-is-the-fraction-of-work-per-second-by-converted-into-heat&show=3391#a3391
<p>Your equation #1 is $$(R+2\lambda x)i=B\ell v=B\ell \frac{dx}{dt}$$ Solving for $x$ as a function of time, and assuming $x(0)=0$, we get $$R+2\lambda x(t)=R e^{kt}$$ where $k=\frac{2\lambda i}{B\ell}$. The velocity of the bar is $v(t)=v_0e^{kt}$ where the initial velocity is $v_0=\frac{iR}{B\ell}$.</p>
<p>We do not need to find an expression for $F$. And we do not need to take account of a resistive force $Bi\ell$ which the magnetic field exerts on the wire. This is because the motion of the wire in the magnetic field is the cause of the current. The magnetic field does not create a current then exert a force on the same current which it has created. If the current had been generated independently of the magnetic field, <em>then</em> there would be a magnetic force on that current.</p>
<p>No work is done by the magnetic field. Only force $F$ does work. The rate of work $P$ done by force $F$ is the sum of 3 components : <br>
(1) the rate $K$ at which the kinetic energy of the bar is increasing;<br>
(2) the power $Q$ dissipated as heat in the resistors; and<br>
(3) the rate $M$ at which energy is being stored in the magnetic field created by the rectangular loop of current. </p>
<p>Now $M$ depends on the self-inductance $L(x)$ of the rectangular loop, which depends on $x$. Even for such a simple geometry an expression for $L(x)$ is difficult to obtain - see <a rel="nofollow" href="https://www.allaboutcircuits.com/tools/rectangle-loop-inductance-calculator/">Rectangle Loop Inductance Calculator</a> in All About Circuits website. I shall assume without justification that $M \ll P$ is negligible.</p>
<p>$$K=\frac{d}{dt}(\frac12mv^2)=mv\frac{dv}{dt}=mkv^2=mkv_0^2 e^{2kt}$$ $$Q=i^2(R+2\lambda x)=i^2Re^{kt}$$ </p>
<p>The fraction of work done by $F$ which is converted into heat is $$\frac{Q}{P}=\frac{Q}{Q+K}=\frac{1}{1+K/Q}=\frac{1}{1+he^{kt}}$$ where $$h=\frac{mkv_0^2}{i^2 R}=\frac{2\lambda m i R}{(B\ell)^3}$$ This is not constant, it decreases with time. The initial fraction at $t=0$ is $\frac{1}{1+h}$. If the rails have zero resistance ($\lambda=0$) then $k=K=h=0$ so $Q=P$ - ie all of the work done by $F$ is dissipated as heat in resistor $R$, as obtained in <a rel="nofollow" href="http://pleclair.ua.edu/ph102/Exams/Sum_2008/ph102_sum08_exam2_SOLN.pdf">Q4 in this worksheet</a>.</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3373/what-is-the-fraction-of-work-per-second-by-converted-into-heat&show=3391#a3391Fri, 22 Feb 2019 17:48:17 +0000Answered: Find minimum distance between dipoles
http://physics.qandaexchange.com/?qa=3372/find-minimum-distance-between-dipoles&show=3390#a3390
<p>Your error is near the start of the calculation. The distance between diagonally opposite charged masses is $\sqrt{\ell^2+d^2}$ not $\sqrt{\ell^2+(\frac{d}{2})^2}$.</p>
<p>Therefore in your answer $d$ should be replaced by $2d$, giving $n=6$.</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3372/find-minimum-distance-between-dipoles&show=3390#a3390Wed, 20 Feb 2019 21:03:31 +0000Find position and velocity at t=0?
http://physics.qandaexchange.com/?qa=3377/find-position-and-velocity-at-t-0
<p>Full question: </p>
<blockquote><p>An object moves with constant acceleration. At t= 2.50 s, the position of the object is x=2.00 m and its velocity is v= 4.50 m/s. At t= 7.00 s, v= -12.0 m/s. <br>
Find: <br>
(a) the position and the velocity at t= 0;<br>
(b) the average speed from 2.50s to 7.00 s, and <br>
(c) the average velocity from 2.50s to 7.00 s.</p>
</blockquote>
<p>I tried using the kinematic equations of motion for constant acceleration but my answers make no sense. </p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3377/find-position-and-velocity-at-t-0Tue, 05 Feb 2019 00:03:55 +0000Answered: Friction between two disks
http://physics.qandaexchange.com/?qa=616/friction-between-two-disks&show=3371#a3371
<p>acceleration of both the disc must be same since there is pure rooling <br>
and friction on smaller disc will be downward and on the bigger dics it will be upwards(Newton's 3rd law) and both will be equal in magnitude.<br>
therefore, writting the equation for bigger dics, f(2R)=2R²/2*(alpha)</p>
<pre><code> =>, f=2N
</code></pre>
<p>since the smaller dics has friction (f) as well as a external force due to which it is accelerating with 2m/s² and the external force was unknown initially. But now it also be found by writing the equation for smaller dics and putting f=2N.</p>
<pre><code> "sorry my previous answer was wrong"
</code></pre>
Physics Problemshttp://physics.qandaexchange.com/?qa=616/friction-between-two-disks&show=3371#a3371Fri, 01 Feb 2019 07:41:22 +0000Frequency modes of the rectangular shell
http://physics.qandaexchange.com/?qa=3357/frequency-modes-of-the-rectangular-shell
<p>This is task i received from my professor:</p>
<blockquote><p>The shapes of three natural modes having the frequencies $\omega_1, \omega_2, \omega_3$ of the rectangular shell are presented in the figure. The exciting pressure $p(t)$ applied uniformly all over the one side of the shell has the form $p(t) = Pe^{jωt}$. <br>
Make a sketch of the normal displacement of the gravity point of the shell against frequency, if the excitation frequency varies within bounds $0.5\omega_1< \omega <2\omega_1$ and static displacement of that point equals to $u_0$.</p>
</blockquote>
<p>Links to photo of frequency nodes (sorry for low quality)-><a rel="nofollow" href="https://ibb.co/b1nh3j9">1</a> .</p>
<p>Can somebody help me and tell me how i should get started?</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3357/frequency-modes-of-the-rectangular-shellMon, 14 Jan 2019 21:42:44 +0000Answered: tension in wires from which a block is suspended
http://physics.qandaexchange.com/?qa=3349/tension-in-wires-from-which-a-block-is-suspended&show=3354#a3354
<p>In order to compare $T$ and $W$, one needs to express one in terms of the other. As explained in the diagram itself, since the block is in translational equilibrium, we have:</p>
<p>$$\vec{T_1}+\vec{T_2}=\vec{W_b}$$</p>
<p>Projecting this relation onto the $Oy$ axis, the components of the tensions are both $T\sin\theta$ (because the angle between $T$ and $Oy$ is $\frac{\pi}{2}-\theta$ and the angle between $T$ and $Ox$ is $\theta$). Therefore:</p>
<p>$$2T\sin\theta=W\implies T=\dfrac{W}{2\sin\theta}$$</p>
<p>The information <em>the strings are almost horizontal</em> is equivalent to $\theta\cong 0$. For very small angles, the sine function is as well extremely small, and almost equal to the angle $\theta$ itself. Take a look at the following examples (I've included them here solely to make sure that you understand why $\sin\theta\to 0$ as $\theta\to 0$)</p>
<ul>
<li>$\sin(0.01)\cong 0.00999983333333\cong 0.01$</li>
<li>$\sin(0.0001) \cong 0.00009999999983\cong 0.0001$</li>
<li>$\sin(0)=0$</li>
</ul>
<p>So the denominator $2\sin\theta\cong0$ and therefore $T\to\infty$. As $W=mg$, clearly $W$ is a finite value (making the reasonable assumption that the mass is not inifite), and therefore <em>clearly</em> $T>W$. </p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3349/tension-in-wires-from-which-a-block-is-suspended&show=3354#a3354Fri, 11 Jan 2019 19:52:11 +0000How to solve this throwing ball off cliff question
http://physics.qandaexchange.com/?qa=3350/how-to-solve-this-throwing-ball-off-cliff-question
<p><img src="https://cdn.pbrd.co/images/HVdO2s0.png" alt=""></p>
<p>For b) I was curious why i couldn't use $s = ut + \frac12at^2$ as my equation as i had all the necessary known values. Instead to get the correct answer i had to use $v=u + at$.</p>
<p>Why is this so? Thank you for your time.</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3350/how-to-solve-this-throwing-ball-off-cliff-questionSun, 06 Jan 2019 15:45:23 +0000Time period of small oscillation- disk and particle system
http://physics.qandaexchange.com/?qa=3346/time-period-of-small-oscillation-disk-and-particle-system
<p><img src="https://cdn.pbrd.co/images/HUWMQA2.jpg" alt=""></p>
<p>Attempt: </p>
<p>$y_{COM} = \dfrac{R}{3} = \dfrac {25}3$</p>
<p>$I = \dfrac 12 (2)R^2 + 1R^2 = 2\times 25 = 50 $</p>
<p>Now, T is given by: $T =2\pi \sqrt{\dfrac{I}{mgl}}$ where l is the distance of the centre of mass from the centre of rotation</p>
<p>So $T =2\pi \sqrt{\dfrac{50}{3 \times 10 \times \dfrac{25}{3}}} = 2\pi\sqrt{\dfrac{1}{5}}$</p>
<p>No option :( </p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3346/time-period-of-small-oscillation-disk-and-particle-systemFri, 04 Jan 2019 22:33:35 +0000Wave optics- number of minimas.
http://physics.qandaexchange.com/?qa=3345/wave-optics-number-of-minimas
<p><img src="https://cdn.pbrd.co/images/HUWujTQ.jpg" alt=""></p>
<p>Attempt: </p>
<p>Let m be the integer associated with $420 nm$<br>
and n be the integer associated with $540 nm$</p>
<p>$d \sin \theta = k \lambda $ $k\in Z$<br>
Clearly, </p>
<p>$m_{max} = 180$ </p>
<p>$n_{max} = 140 $</p>
<p>$(2m+1) \lambda_1 = (2n+1)\lambda_2$ (condition for dark fringes to overlap)</p>
<p>$\implies \dfrac{2m+1}{2n+1} = \dfrac 97$</p>
<p>$\implies m = \dfrac{1+7n +2n }{7}$</p>
<p>Hence, we obtain, for m to be an integer: $2n+1 = 7k$</p>
<p>$\implies n = \dfrac{7k-1}{2}$ where $k \in Z$</p>
<p>Now note that k must be odd since odd-1 = even </p>
<p>Thus, using $n \le 140$, $k_{max} = 39$</p>
<p>Now, we have to consider only odd values of k which are $1,3,...39$ = 20 numbers </p>
<p>Thus, we have 20 minimas on the upper side and 20 on the lower,<br>
Total minimas = $20+20 = 40$</p>
<p>But answer is $D$</p>
<p><strong>Question 1:</strong> What is wrong with my method? </p>
<p><strong>Question 2:</strong> Considering that this is a JEE Mains problem, what is the fastest 2 minute way to do it? </p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3345/wave-optics-number-of-minimasFri, 04 Jan 2019 22:01:11 +0000Answered: oblique collision of a rigid rod with a plane surface
http://physics.qandaexchange.com/?qa=3336/oblique-collision-of-a-rigid-rod-with-a-plane-surface&show=3344#a3344
<p>The normal contact impulse $J$ between the rod and plane has two effects : it changes the velocity of the CM of the rod, and it makes the rod rotate about its CM. </p>
<p>The impulsive force is related to the change in momentum of the CM of the rod, and the impulsive torque to the change in angular momentum of the rod, as follows : $$J=m(v_0+v_1)$$ $$J\frac{L}{2}\cos\theta=I\omega$$ Here $v_0, v_1$ are the vertical velocities of the CM before and after the collision, and $I=\frac{1}{12}mL^2$ is the moment of inertia of the rod about its CM. </p>
<p>The rod is not rotating initially. After the collision, the velocity of the end A which collided with the ground is $v_2'=\frac{L}{2}\omega$ relative to the CM of the rod, directed perpendicular to the rod, ie at an angle of $\theta=60^{\circ}$ to the vertical. The CM of the rod is moving upwards with speed $v_1$ so the vertical component of the velocity of A relative to the ground after collision is $$v_2=v_2'\cos\theta+v_1$$ </p>
<p>The elasticity of the collision is known ($e=1$) so the <strong>Law of Restitution</strong> can be applied to the relative velocities of approach and separation at the point of contact : $$v_2=ev_0$$</p>
<p>From the above equations you can eliminate $v_2, v_2', \omega$ to find the rebound speed $v_1$ of the CM of the rod and thereby the height to which the CM rises after collision.</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3336/oblique-collision-of-a-rigid-rod-with-a-plane-surface&show=3344#a3344Wed, 02 Jan 2019 22:07:06 +0000Answered: Diffusion with reflecting boundary
http://physics.qandaexchange.com/?qa=3254/diffusion-with-reflecting-boundary&show=3343#a3343
<p>You have not gone wrong. You simply have not gone far enough.</p>
<p>You have obtained a quartic equation $$z^4-2z+1=0$$ This has 4 solutions, not only the obvious one $z=1$. The solution $z=1$ requires $t=+\infty$ so this is not the solution you require. You need to find some more solutions. Then check which one describes the situation you are trying to find.</p>
<p>In fact there is one other real solution and 2 imaginary solutions. Probably you should solve numerically (this is a physics question, not maths). </p>
<p>One simple method is to make a first guess $z<em>0$ then reiterate $$z</em>{n+1}=\frac12 (z_n^4+1)$$ With an initial guess of $z_0=0.5$ I get the following iterations :<br>
0.5<br>
0.53125<br>
0.5398259163<br>
0.5424604827<br>
0.543295467<br>
0.543562654<br>
0.5436484118<br>
0.543675964<br>
0.5436848186<br>
0.5436876646</p>
<p>These converge to 0.5437, though not quickly. </p>
<p>Alternatively apply the <a rel="nofollow" href="http://www.sosmath.com/calculus/diff/der07/der07.html">Newton-Raphson Method</a>. Using $$f(z)=z^4-2z+1=0$$$$f'(z)=4z^3-2$$ $$z_{n+1}=z_n-\frac{f(z_n)}{f'(z_n)}$$ again with $z_0=0.5$ we get<br>
0.5<br>
0.5416666667<br>
0.5436837222<br>
0.5436890127<br>
0.5436890127</p>
<p>which converges much more rapidly.</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3254/diffusion-with-reflecting-boundary&show=3343#a3343Wed, 02 Jan 2019 16:42:47 +0000