The ratio of electric potentials at the point | vedclass

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(D) The electric potential $V$ at a point due to a system of point charges is given by $V = \sum \frac{kq_i}{r_i}$.For point $E$,which is the midpoint

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(D) The electric potential $V$ at a point due to a system of point charges is given by $V = \sum \frac{kq_i}{r_i}$.For point $E$,which is the midpoint of side $BC$,the distances are $AE = \sqrt{(2a)^2 + a^2} = a\sqrt{5}$,$BE = a$,$CE = a$,and $DE = \sqrt{(2a)^2 + a^2} = a\sqrt{5}$.The potential at $E$ is $V_E = \frac{1}{4\pi\varepsilon_0} \left( \frac{q}{a\sqrt{5}} + \frac{q}{a} - \frac{q}{a} - \frac{q}{a\sqrt{5}} ight) = 0$.Since $V_E = 0$,the ratio of the potential at $E$ to that at $F$ is $\frac{V_E}{V_F} = \frac{0}{V_F} = 0$ (assuming $V_F 
eq 0$).

The ratio of electric potentials at the point $E$ to that at the point $F$ is

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