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Three charges +q, –q and –q are kept at the vertices of an equilateral triangle of 10 cm side. The potential at the mid point in between –q, –q, if q = 5µC.

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10³V

1 V

–12.8 × 10⁵ V

10 V

answer is C.

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Detailed Solution

V_D = V₁ + V₂ + V₃
$V_{D} = 9 \times 10^{9} [\frac{- 1}{5} + \frac{- 1}{5} + \frac{1}{5 \sqrt{3}}] \times \frac{5 \times 10^{- 6}}{10^{- 2}}$
$V_{D} = - 12.8 \times 10^{5} V$

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