Find the potential difference between the points E and F in the figure given below. Assume E and $F$

Question

Find the potential difference between the points E and F in the figure given below. Assume E and $F$ are the mid point of $A B$ and $C$ respectively\

Professor PC Thomas Classes · Answer

$\mathrm{De}=\mathrm{CE}=\sqrt{(\mathrm{AE})^{2}+(\mathrm{AD})^{2}}=\sqrt{(3)^{2}+(4)^{2}}=5 \mathrm{~m}$ Potential at E.\ $\mathrm{V}_{\mathrm{E}}=\frac{1}{4 \pi \varepsilon_{0}}\left[\frac{\mathrm{q}}{\mathrm{AE}}+\frac{\mathrm{q}}{\mathrm{BE}}+\frac{\mathrm{q} / 2}{\mathrm{DE}}+\frac{\mathrm{q} / 2}{\mathrm{CE}}ight]$\ $=\frac{1}{4 \pi \varepsilon_{0}}\left[\frac{\mathrm{q}}{3}+\frac{\mathrm{q}}{3}+\frac{\mathrm{q} / 2}{5}+\frac{\mathrm{q} / 2}{5}ight]=\frac{1}{4 \pi \varepsilon_{0}}\left[\frac{2 \mathrm{q}}{3}+\frac{\mathrm{q}}{5}ight]$\ $\mathrm{AF}=\mathrm{BF}=\sqrt{(\mathrm{AD})^{2}+(\mathrm{DF})^{2}}=\sqrt{(4)^{2}+(3)^{2}}=5 \mathrm{~m}$\ Potential at F,\ $\mathrm{V}_{\mathrm{F}}=\frac{1}{4 \pi \varepsilon_{0}}\left[\frac{\mathrm{q}}{\mathrm{AF}}+\frac{\mathrm{q}}{\mathrm{BF}}+\frac{\mathrm{q} / 2}{\mathrm{DF}}+\frac{\mathrm{q} / 2}{\mathrm{CF}}ight]$\ $=\frac{1}{4 \pi \varepsilon_{0}}\left[\frac{\mathrm{q}}{5}+\frac{\mathrm{q} / 2}{5}+\frac{\mathrm{q} / 2}{3}ight]=\frac{1}{4 \pi \varepsilon_{0}}\left[\frac{2 \mathrm{q}}{5}+\frac{\mathrm{q}}{3}ight]$\ $	herefore \mathrm{V}_{\mathrm{E}}-\mathrm{V}_{\mathrm{B}}=\frac{1}{4 \pi \varepsilon_{0}}\left[\frac{2 \mathrm{q}}{3}+\frac{\mathrm{q}}{5}-\frac{2 \mathrm{q}}{5}-\frac{\mathrm{q}}{3}ight]$\ $=\frac{\mathrm{q}}{4 \pi \varepsilon_{0}}\left[\frac{1}{3}-\frac{1}{5}ight]$\