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

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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

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$\frac{1}{4 {πε}_{0}} . \frac{2 q}{\sqrt{a^{2} + b^{2}}}$

$\frac{1}{4 {πε}_{0}} . \frac{q}{\sqrt{a^{2} + b^{2}}}$

$\frac{1}{4 {πε}_{0}} . \frac{(- q)}{\sqrt{a^{2} + b^{2}}}$

Zero

answer is B.

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Detailed Solution

Potential at A = Potential due to (+q) charge

+ Potential due to (– q) charge

$= \frac{1}{4 {πε}_{0}} . \frac{q}{\sqrt{a^{2} + b^{2}}} + \frac{1}{4 {πε}_{0}} \frac{(- q)}{\sqrt{a^{2} + b^{2}}} = 0$

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