Question

The potential energy of the infinite system of charges as shown in figure will be


$[$ Consider series expansion of $ln2=(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\dots )]$

• A. $\text{Zero}$
• B. $\frac{-2k{q}^2}{d}ln2$
• C. $\frac{k{q}^2}{d}ln2$
• D. $\frac{-k{q}^2}{d}ln2$

Answer 1

The correct option is B $\frac{-2k{q}^2}{d}ln2$

The given arrangement is a system of infinite charge. We can choose any one charge as our reference point and start calculating potential energy.

Let our reference charge is at $O$.

The electric potential energy due to all the charges located at right-hand side of $O$ will be,
$U=\frac{-q^2}{4\pi {ϵ}_{0}d}+\frac{q^2}{4\pi {ϵ}_{0}2d}+\frac{-q^2}{4\pi {ϵ}_{0}3d}...$

$\Rightarrow U=\frac{-q^2}{4\pi {ϵ}_{0}d}\{1-\frac{1}{2}+\frac{1}{3}\dots \}$
$\Rightarrow U=\frac{-q^2}{4\pi {ϵ}_{0}d}ln2$
$\Rightarrow U=\frac{-k{q}^2}{d}ln2$

Now, the same number of charges are located at left-hand side of the $O$.

So, total potential energy becomes,
$U_{\text{total}}=2U=2\times \frac{-k{q}^2}{d}ln2$
$\Rightarrow U_{\text{total}}=\frac{-2k{q}^2}{d}ln2$

Hence, option (c) is correct.