Question

In the rectangle shown below, the two corners have charges $q_{1}=-5\, \mu C$ and $q_{2}=+2.0\, \mu C .$ The work done in moving a charge $+3.0\, \mu C$ from $B$ to $A$ is (take $\left.1 / 4 \pi \varepsilon_{0}=10^{10} N - m ^{2} / C ^{2}\right)$

• A. 2.8 J
• B. 3.5 J
• C. 4.5 J
• D. 5.5 J

Answer 1

Answer: (A)

Work done $W=3 \times 10^{-6}\left(V_{A}-V_{B}\right)$;
where $V_{A}=10^{10}\left[\frac{\left(-5 \times 10^{-6}\right)}{15 \times 10^{-2}}+\frac{2 \times 10^{-6}}{5 \times 10^{-2}}\right]$
$=\frac{1}{15} \times 10^{6}$ volt
and $V_{B}=10^{10}\left[\frac{\left(2 \times 10^{-6}\right)}{15 \times 10^{-2}}-\frac{5 \times 10^{-6}}{5 \times 10^{-2}}\right]$
$=-\frac{13}{15} \times 10^{6}$ volt
$\therefore W=3 \times 10^{-6}\left[\frac{1}{15} \times 10^{6}-\left(-\frac{13}{15} \times 10^{6}\right)\right]$
$=2.8\, J$