A point charge of $10^{-8} \mathrm{C}$ is placed at origin. The work done in moving a point charge $2 \mu \mathrm{C}$ from point $A(4,4,2) \mathrm{m}$ to $B(2,2,1) \mathrm{m}$ is $\mathrm{J} .\left(\frac{1}{4 \pi \epsilon_{\mathrm{o}}}=9 \times 10^9\right.$ in SI units)
Select the correct option:
A
$45 \times 10^{-6}$
B
00
C
$30 \times 10^{-6}$
D
$15 \times 10^{-6}$
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
Work done by external agent :
$$
\mathrm{W}_{\mathrm{ext}}=\Delta \mathrm{U} ;
$$
$\Delta \mathrm{U} \rightarrow$ Change in potential energy in taking the charge from initial to final configuration
$$
\Rightarrow \quad W_{e x t}=\frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r_f}-\frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r_i}
$$Now, $r_f=\sqrt{(2-0)^2+(2-0)^2+(1-0)^2}=3 \mathrm{~m}$
$$
\begin{aligned}
r_{\mathrm{i}} & =\sqrt{(4-0)^2+(4-0)^2+(2-0)^2}=6 \mathrm{~m} \\
\therefore \quad W_{\text {ext }} & =\left(9 \times 10^9\right) \times\left(10^{-8} \times 2 \times 10^{-6}\right)\left[\frac{1}{3}-\frac{1}{6}\right] \\
& =3 \times 10^{-5} \\
& =30 \times 10^{-6} \mathrm{~J}
\end{aligned}
$$
Correct Option (3)
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