Question

An infinite wire having linear charge density $+\lambda$ is arranged as shown. A charge particle of mass $m$ and charge $q$ is released from point $\text{P}$. Find the initial acceleration of the particle $($at $t=0)$ just after the particle is released.
(Assume the straight parts of wire perpendicular to XY plane.)

• A. $\frac{q\lambda }{2\pi {ϵ}_{0}mR}(-\hat{j})$
• B. $\frac{2q\lambda }{\pi {ϵ}_{0}mR}\left(\hat{k}\right)$
• C. $\frac{q\lambda }{2\pi {ϵ}_{0}mR}\hat{j}$
• D. $\frac{q\lambda }{\pi {ϵ}_{0}mR}(-\hat{j})$

Answer 1

The correct option is A $\frac{q\lambda }{2\pi {ϵ}_{0}mR}(-\hat{j})$

Electric field at point $\text{P}$ due to straight wires are equal and opposite in diection thus they will cancel out.

The net electric field at point $\text{P}$ will be due to semicircular wire only.

Hence,
${\overrightarrow{E}}_{\text{net}}={\overrightarrow{E}}_{\text{semicircular}}=\frac{\lambda }{2\pi {ϵ}_{0}R}(-\hat{j})$

From Newton's second law,
acceleration of the particle at point $\text{P}$,
$$\overrightarrow{a}=\frac{q{\overrightarrow{E}}_{\text{net}}}{m}$$
Substituting the value of ${\overrightarrow{E}}_{\text{net}}$, we get

$a=\frac{q\lambda }{2\pi {ϵ}_{0}mR}(-\hat{j})$

Hence, option (d) is correct answer.