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How to apply the gauss law when charge density is a function of not only $r$?

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The problem goes as

A sphere of radius R, centre at origin carries charges with density $\rho(r,\theta)= \frac{KR(R-2r)sin\theta}{r^2}$ where K is a constant and $r,\theta$ are usual spherical co-ordinates. Find the potential and field for points on the $z$ axis far from the centre.

Now in that question how can I apply Gauss' Law because the equipotential surfaces would not be sphere anymore?

asked Mar 17 in Physics Problems by Nobody recognizable1 (130 points)
edited Mar 27 by sammy gerbil
Gauss' Law applies. However, it is not useful here because the charge distribution is not spherically symmetric, as you realise. Gauss' Law is useful when the potential is constant over the chosen surface, or varies in a known simple manner.

For this problem you will have to use integration.

I would divide the sphere into rings of radius $s=r\sin\theta$ and thickness $dz, ds$ parallel and perpendicular to the $z$ axis respectively. Each ring is the same distance $r$ from the centre and has the same azimuth angle $\theta$ so it has constant charge density. However, note that although all rings are centred on the $z$ axis, not all have their centre coinciding with the centre of the sphere. You will need to integrate from $s=0 \to \sqrt{R^2-z^2}$ and from $z=-R \to +R$. Note that $r^2=z^2+s^2$ and $\sin\theta=s/r$.

You may also have to make an approximation at some stage, as suggested by the words "far from the centre" - ie where $z \gg R$. This means that if you expand your answer in powers of $z$ then you keep only the smallest power of $z$.
I see and what about the shell theorem could i say $E(r) = Q/4\pi \epsilon_0 r^2$ as the distance is fairly large so that it could be considered as a point charge?
Yes good idea you can do that, but it requires that the overall charge $Q$ on the sphere is non-zero. In this case the inner part of the sphere $r<R/2$ is +ve and the outer part $r>R/2$ is -ve. So you need to find out if these two charges cancel out or if there is an overall non-zero charge.
@sammygerbil hi professor i think total charge is zero . As $\int\int\int \rho r^2sin\theta dr d\theta d\phi$ over appropriate limit gives zero. So is the shell theorem not applicable here?
@Nobody recognised  If total charge is zero then the electric field of the charged sphere will not be the same as that of a point charge at large distances ($E\propto 1/r^2$). It will (usually) behave as an electric dipole ($E\propto 1/r^3$) or maybe a quadrupole ($E \propto 1/r^4$) or octopole ($E \propto 1/r^5$). See  http://physics.qandaexchange.com/?qa=3393/electric-field-far-away-of-a-bunch-of-charges.

I think you will have to integrate to get the answer.

1 Answer

1 vote

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 multipole expansion : $$V(z)=\frac{A}{r}+\frac{B}{r^2}+\frac{C}{r^3}+\frac{D}{r^4}+...$$ 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 Electric field far away from a bunch of charges.

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.

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,

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 $$dQ=2\pi y \rho r d\theta dr=2\pi\rho r^2\sin\theta d\theta dr$$

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 $$dV=\frac{dQ}{4\pi \epsilon_0 }(\frac{1}{s_1}+\frac{1}{s_2})=\frac{KR(R-2r)}{2\epsilon_0 z}(\frac{1}{\sqrt{1- x}}+\frac{1}{\sqrt{1+ x}})\sin^2\theta d\theta dr$$

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 $$\frac12 (\frac{1}{\sqrt{1-x}}+\frac{1}{\sqrt{1+x}})=1+\frac38x^2+\frac{35}{128}x^4++\frac{231}{1024}x^6+...$$

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 $\theta=0 \to \pi/2$.

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 $$x^2=p^2(c-p)^2\approx \frac{r^2}{z^2}\cos^2\theta$$

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 $$E_z=-\frac{dV}{dz}=-\frac{3\pi KR^5}{256\epsilon_0 z^4}$$

Notes :

  1. We should expect the potential to be -ve because the outer part of the sphere (which is closest to P) is negatively charged.

  2. The dominant term in the charge distribution is not a dipole $V\propto 1/r^2$ but a quadrupole $V\propto 1/r^3$. We should expect to lose the dipole term because the centres of +ve and -ve charge coincide.

  3. This result gives only the leading term in the multipole expansion 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 $z \gt R$.

answered Mar 26 by sammy gerbil (28,746 points)
edited Mar 27 by sammy gerbil
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