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Q.

Four identical charges each +Q are at the vertices of a square of side “L”. The electric potential at the center of the square is ‘V’. If one of the charges is reversed in sign, the potential now at the center of the square will be

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a

$V/2$

b

$- V/2$

c

$-3V/4$

d

$3V/4$

answer is C.

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Detailed Solution

$V_{1} = K \times \frac{4 Q}{r}$

$V_{2} = K \times \frac{2 Q}{r}$

$\frac{V_{1}}{V_{2}} = \frac{2}{1}$

$V_{2} = 2 V_{1}$

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Four identical charges each +Q are at the vertices of a square of side “L”. The electric potential at the center of the square is ‘V’. If one of the charges is reversed in sign, the potential now at the center of the square will be