Question

Find the potential difference between the points $E$ and $F$ in the figure given below. Assume $E$ and $F$ are the midpoints of $AB$ and $DC$ 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)

$DE=CE=\sqrt{(AE{)}^2+(AD{)}^2}$
$=\sqrt{(3{)}^2+(4{)}^2}=5m$
Potential at $E$,
$V_E=\frac{1}{4\pi {\epsilon }_0}\left[\frac{q}{AE}+\frac{q}{BE}+\frac{q/2}{DE}+\frac{q/2}{CE}\right]$
$=\frac{1}{4\pi {\epsilon }_0}\left[\frac{q}{3}+\frac{q}{3}+\frac{q/2}{5}+\frac{q/2}{5}\right]=\frac{1}{4\pi {\epsilon }_0}\left[\frac{2q}{3}+\frac{q}{5}\right]$
$AF=BF=\sqrt{(AD{)}^2+(DF{)}^2}$
$=\sqrt{(4{)}^2+(3{)}^2}=5m$
Potential at $F$,
$V_F=\frac{1}{4\pi {\epsilon }_0}\left[\frac{q}{AF}+\frac{q}{BF}+\frac{q/2}{DF}+\frac{q/2}{CF}\right]$
$=\frac{1}{4\pi {\epsilon }_0}\left[\frac{q}{5}+\frac{q}{5}+\frac{q/2}{3}+\frac{q/2}{3}\right]=\frac{1}{4\pi {\epsilon }_0}\left[\frac{2q}{5}+\frac{q}{3}\right]$
$\therefore V_E-V_F$
$=\frac{1}{4\pi {\epsilon }_0}\left[\frac{2q}{3}+\frac{q}{5}-\frac{2q}{5}-\frac{q}{3}\right]=\frac{q}{4\pi {\epsilon }_0}\left[\frac{1}{3}-\frac{1}{5}\right]$
$=\frac{2q}{15\times 4\pi {\epsilon }_0}$
$=\frac{2}{15}\times 9\times 10^9\times q$
$=1.2\times 10^{9}q$ volt.