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# Charges +q and -q are placed on vertices B and C of isosceles triangle. The potential at vertex A is :

$(A)\;\frac{1}{4 \pi \in_0} \frac{2q}{\sqrt {a^2+b^2}} \\ (B)\;zero \\ (C)\; \frac{1}{4 \pi \in_0} \frac{q}{\sqrt {a^2+b^2}} \\ (D)\;None$

V at A due to $+q=\large\frac{1}{4 \pi \in _0} \frac{(+q)}{\sqrt {a^2 +b^2}}$
V at A due to $-q=\large\frac{1}{4 \pi \in _0} \frac{(-q)}{\sqrt {a^2 +b^2}}$
Total $V= V_{+q} +V_{-q}$
$\qquad=0$
Hence B is the correct answer.