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Electric field vanishes for 3 charges at the corners of an equilateral triangle

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Three identical point charges are placed on the vertices of an equilateral triangle . At how many places within the triangle electric field vanishes .

These are arbitrary lines .
But this is going to be very hard to solve these equations .
Is there any other way .

asked Apr 14 in Physics Problems by koolman (4,116 points)
edited May 23 by sammy gerbil
I'm going to guess 7 points but awaiting an answer.  This is based on 1 in the middle, 3 close to the sides, and 3 close to the points.
I think 3 close to the points should not be in the answer as there would be very high electric field due to that point
And I think you are just checking on the median .
See "Electric Field inside a regular polygon with corner charges", https://physics.stackexchange.com/q/108929

1 Answer

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Best answer

The null points where electric field is zero must lie inside the triangle and on the axes of reflection symmetry.

Consider a triangle with side $2a$. Suppose a null point P lies on the axis through charge 3, as in the diagram below.

Because of symmetry, the horizontal components of the electric fields at P due to charges 1 & 2 cancel out. The vertical components of electric field for the 3 charges are
$E_1=\frac{K}{r_1^2}\sin\theta=\frac{Ky}{r_1^3}$
$E_2=E_1$
$E_3=\frac{K}{r_3^2}$

Using geometry we also have
$r_1^2=a^2+y^2$
$r_3=h-y$
$h=a\sqrt3$

For P to be a point where the total electric field is zero, we must have $E_1+E_2=E_3$. After substituting from the above equations and rearranging we get an expression containing the single variable $y$ and parameter $a$:
$2y(a\sqrt3-y)^2=(a^2+y^2)^{3/2}$

We wish to find the proportional distance of P from vertex 3, so let $h=1$. The equation becomes
$2y(1-y)^2=(\frac13+y^2)^{3/2}$
If you wish you can expand this into a polynomial equation of degree 6. It may be possible to obtain an analytic solution, but this will be extremely difficult. It is much easier to solve the equation numerically (eg using WolframAlpha ). The solutions are
$y\approx 0.143521$
$y\approx 3.58216$
$y=\frac13 = 0.333333...$

We must reject the 2nd solution which lies outside the triangle and is unphysical. The 3rd solution is the trivial one in which P lies at the centroid, ie $\frac13h$ from each side. The 1st solution is non-trivial : P lies approx $\frac17 h$ from the nearest side.

There are 3 non-trivial positions for P (one on each median) and 1 trivial position, making a total of 4 points at which the electric field is zero.

answered Apr 15 by sammy gerbil (26,166 points)
selected Apr 22 by koolman
How can we say it is on the axes of reflection symmetry.
Good question, which is not easy for me to answer, because I made that statement by intuition without having a logical reason. I shall try to think of one.  Symmetry is required, but perhaps this should be rotational symmetry.
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