Question

Two point charges $100\mu \text{C}$ and $5\mu \text{C}$ are placed at points $\text{A}$ and $\text{B}$ respectively with $AB=40\text{cm}$.The workdone by external force in displacing the charge $5\mu \text{C}$ from $\text{B}$ to $\text{C}$ is
$\text{Here}BC=30\text{cm},$ $\angle ABC=\frac{\pi }{2}$

• A. $\frac{81}{20}\text{J}$
• B. $\frac{-9}{25}\text{J}$
• C. $\frac{-9}{4}\text{J}$
• D. $9\text{J}$

Answer 1

The correct option is C $\frac{-9}{4}\text{J}$




Workdone,

$W=\Delta PE=P{E}_{AC}-P{E}_{AB}$

$P{E}_{AB}=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{r_{AB}}.......\left(1\right)$

$P{E}_{AC}=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{r_{AC}}.......\left(2\right)$

$W=\frac{q_1{q}_2}{4\pi {\epsilon }_0}[\frac{1}{r_{AB}}-\frac{1}{r_{AC}}]$

$=9\times {10}^9\times (100\times {10}^{-6})(5\times {10}^{-6})(\frac{1}{50}-\frac{1}{40})\times {10}^2$

$=90\times 5\times \left[\frac{40-50}{40\times 50}\right]$

$=\frac{-90\times 10\times 5}{40\times 50}$

$=-\frac{9}{4}\text{J}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, (c) is the correct answer.