Question

Two charges $q_1$ and $q_2$ are placed $30\text{cm}$ apart as shown in the figure. A third charge $q_3$ is moved along the arc of a circle of radius $40\text{cm}$ from $C$ to $D$. The change in potential energy of the system is $\frac{x{q}_3}{4\pi {ϵ}_0}$ where $x$ is

• A. $8q_2$
  
• B. $6q_2$
• C. $8q_1$
• D. $6q_1$

Answer 1

The correct option is A $8q_2$


Given that,
$AB=30\text{cm}=0.3\text{m}$
$AC=40\text{cm}=0.4\text{m}$
Initially, $q_3$ is at point $C$.
$BC=\sqrt{A{B}^2+A{C}^2}$
$\Rightarrow BC=\sqrt{{0.3}^2+{0.4}^2}$
$\Rightarrow BC=0.5\text{m}$

The initial potential energy of the system when charge $q_3$ was at $C$,
$U_i=k\{\frac{q_1{q}_2}{AB}+\frac{q_1{q}_3}{AC}+\frac{q_2{q}_3}{BC}\}$
$\Rightarrow U_i=\frac{1}{4\pi {ϵ}_0}\{\frac{q_1{q}_2}{0.3}+\frac{q_1{q}_3}{0.4}+\frac{q_2{q}_3}{0.5}\}$


Now, final potential energy of the system when charge $q_3$ is moved from $C$ to $D$,
$U_f=\frac{1}{4\pi {ϵ}_0}\{\frac{q_1{q}_3}{0.4}+\frac{q_1{q}_2}{0.3}+\frac{q_2{q}_3}{0.1}\}$

So, $U_f-U_i=\frac{1}{4\pi {\epsilon }_0}\{\frac{q_2{q}_3}{0.1}-\frac{q_2{q}_3}{0.5}\}$

$\Rightarrow U_f-U_i=\frac{10}{4\pi {\epsilon }_0}\{\frac{q_2{q}_3}{1}-\frac{q_2{q}_3}{5}\}$

$\Rightarrow U_f-U_i=\frac{10}{4\pi {ϵ}_o}\times \frac{4q_2{q}_3}{5}$

$\Rightarrow U_f-U_i=\frac{8q_2{q}_3}{4\pi {ϵ}_0}=\frac{x{q}_3}{4\pi {ϵ}_0}$ $[$given$]$

$\Rightarrow x=8q_2$

Hence, option (a) is correct.