Answer:
[tex]q_{3} = - 4.8 \times 10^{-6} C[/tex]
Explanation:
given two charges
[tex]q_{1} = q_{2}= q = +1.6 \times 10^6 C \\[/tex]
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
[tex]V = \frac{kq}{r}[/tex]
thus we have
[tex](V_{C})_{initial} = \frac{kq}{r}+\frac{kq}{r} \\(V_{C})_{initial} = 2\frac{kq}{r}[/tex] ...................(2)
let charge [tex]q_{3}[/tex] is placed at the center of two charge q as shown in figure
from Pythagoras we get
[tex]d = \sqrt{( r^2+r^2)}[/tex]
[tex]\frac{d}{2} = \frac{r}{\sqrt{2}}[/tex]
we get
[tex](V_{c})_{final} = 2\frac{kq}{r}+\frac{kq_{3}}{d/2}[/tex]
[tex](V_{c})_{final} = 2\frac{kq}{r}+\frac{k\sqrt{2} q_{3}}{r}[/tex] ........(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
[tex](V_{C})_{initial} = -(V_{C})_{final}[/tex]
[tex]we \ \ get \\\frac{2kq}{r} + \frac{kq_{3}\sqrt{2}}{r} = - \frac{2kq}{r}[/tex]
on solving we get
[tex]q_{3} = - 4.8 \times 10^{-6} C[/tex]
