Question

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

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

Answer 1

The Correct Option is A

Now, potential at $A$ $ V_{A}=\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{B}}{A B}+\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q_{C}}{A C}+\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q_{D}}{A D} $ $= \frac{1}{4 \pi \varepsilon_{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} $ $= 16.5 \times 10^{-1}=1.65 V$