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2. Electric Potential and Capacitance
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The ratio of electric potentials at the point $E$ to that at the point $F$ is
A
$\left( {\frac{{\sqrt 5 - 1}}{{\sqrt 5 }}} \right)$
B
$-\left( {\frac{{\sqrt 5 - 1}}{{\sqrt 5 }}} \right)$
C
$\sqrt 2$
D
Zero
Solution
$\mathrm{V}_{\mathrm{E}}=\frac{\mathrm{q}}{4 \pi \varepsilon_{0}}\left[\frac{1}{\mathrm{AE}}+\frac{-1}{\mathrm{DE}}+\frac{1}{\mathrm{BE}}+\frac{-1}{\mathrm{CE}}\right]$
$\therefore A E=D E$ and $B E=C E$
$\therefore V_{E}=0$
Although $\mathrm{V}_{\mathrm{F}} \neq 0$ but $\frac{\mathrm{V}_{\mathrm{E}}}{\mathrm{V}_{\mathrm{F}}}=\frac{\mathrm{O}}{\mathrm{V}_{\mathrm{F}}}=\mathrm{Zero}$
Standard 12
Physics
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