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As shown in the figure, charges $+\mathrm{q}$ and $-\mathrm{q}$ are placed at the vertices $\mathrm{B}$ and $\mathrm{C}$ of an isosceles triangle. The potential at the vertex $\mathrm{A}$ is
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zero
Potential at A
$=+\frac{q}{\sqrt{a^{2}+b^{2}}}-\frac{q}{\sqrt{a^{2}+b^{2}}}=0$
$=+\frac{q}{\sqrt{a^{2}+b^{2}}}-\frac{q}{\sqrt{a^{2}+b^{2}}}=0$
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