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 $V = \frac{1}{4\pi \epsilon_0} \cdot \frac{Q}{r}$.In

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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 $V = \frac{1}{4\pi \epsilon_0} \cdot \frac{Q}{r}$.In the given isosceles triangle,the distance from vertex $A$ to vertex $B$ is $a$,and the distance from vertex $A$ to vertex $C$ is also $a$ (since it is an isosceles triangle with $AB = AC = a$).The potential at vertex $A$ due to charge $+q$ at $B$ is $V_B = \frac{1}{4\pi \epsilon_0} \cdot \frac{+q}{a}$.The potential at vertex $A$ due to charge $-q$ at $C$ is $V_C = \frac{1}{4\pi \epsilon_0} \cdot \frac{-q}{a}$.The total potential at vertex $A$ is the algebraic sum of the individual potentials: $V_A = V_B + V_C = \frac{1}{4\pi \epsilon_0} \cdot \frac{q}{a} + \frac{1}{4\pi \epsilon_0} \cdot \frac{-q}{a} = 0$.

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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