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A square of side $\sqrt{2} m$ has charges $+ 2 \times 10^{- 9} C, 1 \times 10^{- 9} C$ , $- 2 \times 10^{- 9} C$ and $- 3 \times 10^{- 9} C$
respectively at its corners. Potential at the centre of the square is

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-8V

-18V

+18V

+8V

answer is D.

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Detailed Solution

$V = \frac{K}{r} (q_{1} + q_{2} + q_{3} + q_{4})$

$here r = \frac{a}{\sqrt{2}} if a is the side of the square r is half of the diagonal .$

$Putiing the values V = \frac{9 \times 10^{9}}{1} (2 + 1 - 2 - 3) 10^{- 9} = - 18 V$

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