Question

Identical point charges of +1.6 μC are fixed to diagonally opposite corners of a square. A third charge is then fixed at the center of the square, such that it causes the potentials at the empty corners to change signs without changing magnitudes. Find the sign and magnitude of the third charge.

Answer 1

Answer:

$q_{3} = - 4.8 \times 10^{-6} C$

Explanation:

given two charges

$q_{1} = q_{2}= q = +1.6 \times 10^6 C$

let A , B, C, D are the corner where C and D are empty so we consider potential at  C due to  A and B is

using

$V = \frac{kq}{r}$

thus we have

$(V_{C})_{initial} = \frac{kq}{r}+\frac{kq}{r} \\(V_{C})_{initial} = 2\frac{kq}{r}$   ...................(2)

let  charge $q_{3}$ is placed at the center of two charge q  as shown in figure

from Pythagoras  we get

$d = \sqrt{( r^2+r^2)}$

$\frac{d}{2} = \frac{r}{\sqrt{2}}$

we get

$(V_{c})_{final} = 2\frac{kq}{r}+\frac{kq_{3}}{d/2}$

$(V_{c})_{final} = 2\frac{kq}{r}+\frac{k\sqrt{2} q_{3}}{r}$ ........(3)

according to question such that it causes the potentials at the empty corners to change signs without changing magnitudes. from first and third equation

$(V_{C})_{initial} = -(V_{C})_{final}$

$we \ \ get \\\frac{2kq}{r} + \frac{kq_{3}\sqrt{2}}{r} = - \frac{2kq}{r}$

on solving we get

$q_{3} = - 4.8 \times 10^{-6} C$