Question

Two points charges $4q$ and$–q$ are fixed on the$x-$axis at $x=\frac{-d}{2}$ and $x=\frac{d}{2}$, respectively. If a third point charge $‘q’$is taken from the origin to $x=d$ along the semicircle as shown in the figure, the energy of the charge will:

• A. decrease by $\frac{q^2}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}\mathrm{d}}$
• B. decrease by $\frac{4q^2}{3{\mathrm{\pi \epsilon }}_{\mathrm{o}}\mathrm{d}}$
• C. increases by$\frac{3q^2}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}\mathrm{d}}$
• D. decrease by $\frac{2q^2}{3\mathrm{\pi }{\epsilon }_{\mathrm{o}}\mathrm{d}}$

Answer 1

The correct option is B

decrease by $\frac{4q^2}{3{\mathrm{\pi \epsilon }}_{\mathrm{o}}\mathrm{d}}$

Step 1: Given

1. The value of the two-point charges $4q$ and$–q$ are at $x=\frac{-d}{2}$ and $x=\frac{d}{2}$.
2. The value of the third charge is $q$ at $x=d$.

Step 2: To find

The energy of the third charge

Step 3: Solution

The potential energy $U=\frac{q_1{q}_2}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}\mathrm{R}}$

$$\begin{gathered} q_1=Valueoffirstcharge \\ {q}_2=Valueofsecondcharge \\ R=Dis\tan cebetweenthecharges \end{gathered}$$

Let initial and final potential energy be $U_i$and $U_f$

Potential energy at initial point = potential energy due to $4q$ and the moving charge $q$ at A + potential energy due $-q$ and the moving charge $q$ at A.

$$\begin{gathered} U_i=\frac{1}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}}\left[\frac{4q\times q}{\left({\displaystyle \frac{d}{2}}\right)}+\frac{q\left(-q\right)}{\left({\displaystyle \frac{d}{2}}\right)}\right] \\ {U}_i=\frac{1}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}}\left[\frac{4q^2}{\left({\displaystyle \frac{d}{2}}\right)}-\frac{q^2}{\left({\displaystyle \frac{d}{2}}\right)}\right] \end{gathered}$$

Potential energy at final point = potential energy due $4q$ and the moving charge $q$ at B + potential energy due $-q$ and the moving charge $q$ at B.

$$\begin{gathered} U_f=\frac{1}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}}\left[\frac{4q\times q}{\left({\displaystyle \frac{3d}{2}}\right)}+\frac{q\left(-q\right)}{\left({\displaystyle \frac{d}{2}}\right)}\right] \\ {U}_f=\frac{1}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}}\left[\frac{4q^2}{\left({\displaystyle \frac{3d}{2}}\right)}-\frac{q^2}{\left({\displaystyle \frac{d}{2}}\right)}\right] \end{gathered}$$

The change in the energy

$∆U=U_f-U_i$

$$\begin{gathered} \Rightarrow ∆U=\frac{1}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}}\left[\frac{4q^2}{\left({\displaystyle \frac{3d}{2}}\right)}-\frac{4q^2}{\left({\displaystyle \frac{d}{2}}\right)}\right] \\ \Rightarrow ∆U=\frac{4q^2}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}}\left[\frac{2}{3d}-\frac{2}{d}\right] \\ \Rightarrow ∆U=\frac{-4q^2}{3{\mathrm{\pi \epsilon }}_{\mathrm{o}}d} \end{gathered}$$

Here, the negative sign shows that energy is decreasing.

Hence, the correct option is $B$.