Question

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

• A. $\left(1.2 \times 10^{9} q\right)$ volt
• B. $\left(1.8 \times 10^{9} q\right)$ volt
• C. $\left(1.5 \times 10^{9} q\right)$ volt
• D. $\left(3 \times 10^{9} q\right)$ volt

Answer 1

Answer: (A)

$D E=C E=\sqrt{(A E)^{2}+(A D)^{2}}$
$=\sqrt{(3)^{2}+(4)^{2}}=5 m$
Potential at $E$,
$V_{E}=\frac{1}{4 \pi \varepsilon_{0}}\left[\frac{q}{A E}+\frac{q}{B E}+\frac{q / 2}{D E}+\frac{q / 2}{C E}\right] $
$=\frac{1}{4 \pi \varepsilon_{0}}\left[\frac{q}{3}+\frac{q}{3}+\frac{q / 2}{5}+\frac{q / 2}{5}\right]=\frac{1}{4 \pi \varepsilon_{0}}\left[\frac{2 q}{3}+\frac{q}{5}\right] $
$A F=B F=\sqrt{(A D)^{2}+(D F)^{2}}$
$=\sqrt{(4)^{2}+(3)^{2}}=5 \,m$
Potential at $F$,
$V_{F}=\frac{1}{4 \pi \varepsilon_{0}}\left[\frac{q}{A F}+\frac{q}{B F}+\frac{q / 2}{D F}+\frac{q / 2}{C F}\right]$
$= \frac{1}{4 \pi \varepsilon_{0}}\left[\frac{q}{5}+\frac{q}{5}+\frac{q / 2}{3}+\frac{q / 2}{3}\right]=\frac{1}{4 \pi \varepsilon_{0}}\left[\frac{2 q}{5}+\frac{q}{3}\right] $
$\therefore V_{E}-V_{F} $
$=\frac{1}{4 \pi \varepsilon_{0}}\left[\frac{2 q}{3}+\frac{q}{5}-\frac{2 q}{5}-\frac{q}{3}\right]=\frac{q}{4 \pi \varepsilon_{0}}\left[\frac{1}{3}-\frac{1}{5}\right] $
$=\frac{2 q}{15 \times 4 \pi \varepsilon_{0}}$
$=\frac{2}{15} \times 9 \times 10^{9} \times q$
$=1.2 \times 10^{9} q $ volt.