As shown in the figure,charges +q and -q are | vedclass

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(B) The electric potential $V$ at a point due to a point charge $Q$ at a distance $r$ is given by $V = \frac{1}{4\pi\varepsilon_0} \cdot \frac{Q}{r}$.

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(B) The electric potential $V$ at a point due to a point charge $Q$ at a distance $r$ is given by $V = \frac{1}{4\pi\varepsilon_0} \cdot \frac{Q}{r}$.In the given isosceles triangle,the distance from vertex $A$ to both vertices $B$ and $C$ is $r = \sqrt{a^2 + b^2}$ by the Pythagorean theorem.The potential at vertex $A$ is the algebraic sum of the potentials due to charges at $B$ and $C$:$V_A = V_B + V_C$$V_A = \frac{1}{4\pi\varepsilon_0} \cdot \frac{q}{\sqrt{a^2 + b^2}} + \frac{1}{4\pi\varepsilon_0} \cdot \frac{-q}{\sqrt{a^2 + b^2}}$$V_A = \frac{1}{4\pi\varepsilon_0} \cdot \frac{q - q}{\sqrt{a^2 + b^2}} = 0$.

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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