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Four point charges −Q, −q, 2q and 2Q are placed, one at each corner of the square. The relation between Q and q for which the potential at the centre of the square is zero is
A
$Q = \frac{1}{q}$
B
Q = −q
C
$Q = -\frac{1}{q}$
D
Q = q
Detailed Solution
All four corners are at the same distance x from the centre.
$V = \frac{k(-Q)}{x} + \frac{k(-q)}{x} + \frac{k(2q)}{x} + \frac{k(2Q)}{x} = 0$
$-Q - q + 2q + 2Q = 0 \Rightarrow Q + q = 0$
$Q = -q$
$V = \frac{k(-Q)}{x} + \frac{k(-q)}{x} + \frac{k(2q)}{x} + \frac{k(2Q)}{x} = 0$
$-Q - q + 2q + 2Q = 0 \Rightarrow Q + q = 0$
$Q = -q$