Question

Two charges q₁ and q₂ are placed at (0, 0, d) and (0, 0, – d) respectively. Find locus of points where the potential a zero.

Answer 1

Following the principle of superposition of potentials as
described in last section, let us find the potential V due to a collection of discrete point charges q₁, q₂, …, q_n, at a point P.

The potential at P due to the system of point charges is given as the sum of their individual potentials at P, V = ${\displaystyle \frac{1}{4\pi {\epsilon }_0}\sum \frac{q_i}{r_i}}$

As we know, the potential at point P is V = ${\displaystyle \sum V_i}$,

 Where ${\displaystyle V_i=\frac{q_i}{4\pi {\epsilon }_0};r_i}$ = magnitude of position vector P relative to q_r

Then ${\displaystyle V=\frac{1}{4\pi {\epsilon }_0}\sum \frac{q_i}{r_{\text{pi}}}}$

Let us take a point on the required plane as (x, y, z). The two charges lies on z-axis at a separation of 2d. The potential at the point P due to two charge is given by

${\displaystyle \frac{q_1}{\sqrt{x^2+y^2+{(z-d)}^2}}+\frac{q_2}{\sqrt{x^2+y^2+{(z+d)}^2}}}$ = 0

∴ ${\displaystyle \frac{q_1}{\sqrt{x^2+y^2+{(z-d)}^2}}=\frac{-q_2}{\sqrt{x^2+y^2+{(z+d)}^2}}}$

On squaring and simplifying, we get

${\displaystyle x^2+y^2+z^2+\left[\frac{{\left(\frac{q_1}{q_2}\right)}^2+1}{{\left(\frac{q_1}{q_2}\right)}^2-1}\right]\left(2zd\right)+d^2-0}$

The standard equation of sphere is ${\displaystyle x^2+y^2+z^2+2ux+2uy+2wz+g}$ = 0

With centre ${\displaystyle (-u,-v,-w)}$ and radius ${\displaystyle \sqrt{u^2+v^2+w^2}-g}$

Hence centre of sphere will be ${\displaystyle (0,0-d\left[\frac{q_1^2+q_2^2}{q_1^2-q_2^2}\right])}$

And radius is ${\displaystyle r=\sqrt{{\left(d\left[\frac{q_1^2+q_2^2}{q_1^2-q_2^2}\right]\right)}^2-d^2}=\frac{2q_1{q}_{2}d}{q_1^2-q_2^2}}$