Ask Questions, Get Answers


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 $

1 Answer

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 $V= V_{+q} +V_{-q}$
Hence B is the correct answer.
answered Jan 8, 2014 by meena.p

Related questions

Download clay6 mobile appDownload clay6 mobile app