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Q.

Four point charges -Q,-q,2q and 2Q are placed, one at each corner of the square. The relation between Q and q fo which the potential at the centre of the square is zero, is

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a

$Q = - q$

b

Q = q

c

$Q = - \frac{1}{q}$

d

$Q = \frac{1}{q}$

answer is C.

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Detailed Solution

Let the side of the square be = a Potential at centre
$V_{e} = \frac{K \times - Q}{\frac{a}{\sqrt{2}}} + \frac{K \times - q}{\frac{a}{\sqrt{2}}} + \frac{K \times 2 q}{\frac{a}{\sqrt{2}}} + \frac{K \times 2 q}{\frac{a}{\sqrt{2}}}$
$O = \frac{K Q}{\frac{a}{\sqrt{2}}} + \frac{K q}{\frac{a}{\sqrt{2}}}$
$= Q = - q$

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