As shown in the figure, charges $ + q$ and $ - q$ are placed at the vertices $B$ and $C$ of an isosceles triangle. The potential at the vertex $A$ is
AIIMS 2002, Easy
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Potential at $A$ = Potential due to $(+q)$ charge + Potential due to $(-q)$ charge
$ = \frac{1}{{4\pi {\varepsilon _0}}}.\frac{q}{{\sqrt {{a^2} + {b^2}} }} + \frac{1}{{4\pi {\varepsilon _0}}}\frac{{( - q)}}{{\sqrt {{a^2} + {b^2}} }} = 0$
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