Question

$ABCD$ is a rectangle. At corners $B,C$ and $D$ of the rectangle are placed charges $+10\times 10^{-10}C,-20\times 10^{-12}C$ and $10\times 10^{-12}C$, respectively. Calculate the potential at the fourth corner. (The side $AB=4cm$ and $BC=3cm)$

• A. 1.65 V
• B. 0.165 V
• C. 16.5 V
• D. 2.65 V

Answer 1

Answer: (A)

Now, potential at $A$
$V_A=\frac{1}{4\pi {\epsilon }_0}\frac{q_B}{AB}+\frac{1}{4\pi {\epsilon }_0}\cdot \frac{q_C}{AC}+\frac{1}{4\pi {\epsilon }_0}\cdot \frac{q_D}{AD}$
$=\frac{1}{4\pi {\epsilon }_0}\left[\frac{10\times 10^{-12}}{4\times 10^{-2}}-\frac{20\times 10^{-12}}{5\times 10^{-2}}+\frac{10\times 10^{-12}}{3\times 10^{-2}}\right]$
$=9\times 10^9\times 10^{-10}\left[\frac{10}{4}-\frac{20}{5}+\frac{10}{3}\right]$
$=\frac{9\times 10^{-1}\times 11}{6}$
$=16.5\times 10^{-1}=1.65V$