The ratio of electric potentials at the point $E$ to that at the point $F$ is
Medium
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$\mathrm{V}_{\mathrm{E}}=\frac{\mathrm{q}}{4 \pi \varepsilon_{0}}\left[\frac{1}{\mathrm{AE}}+\frac{-1}{\mathrm{DE}}+\frac{1}{\mathrm{BE}}+\frac{-1}{\mathrm{CE}}\right]$
$\therefore A E=D E$ and $B E=C E$
$\therefore V_{E}=0$
Although $\mathrm{V}_{\mathrm{F}} \neq 0$ but $\frac{\mathrm{V}_{\mathrm{E}}}{\mathrm{V}_{\mathrm{F}}}=\frac{\mathrm{O}}{\mathrm{V}_{\mathrm{F}}}=\mathrm{Zero}$
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Reason : The field just outside the capacitor is $\frac{\sigma }{{{\varepsilon _0}}}$. ( $\sigma $ is the charge density).