Question

Find the force experienced by a semicircular rod having a charge q as shown in figure. Radius of the wire is R, and the line of charge with linear charge density $\lambda$ passes through its center and is perpendicular to the plane of wire.

• A. $\frac{\lambda q}{2{\pi }^2{\epsilon }_{0}R}$
• B. $\frac{\lambda q}{{\pi }^2{\epsilon }_{0}R}$
• C. $\frac{\lambda q}{4{\pi }^2{\epsilon }_{0}R}$
• D. $\frac{\lambda q}{8{\pi }^2{\epsilon }_{0}R}$

Answer 1

The correct option is B $\frac{\lambda q}{{\pi }^2{\epsilon }_{0}R}$

$F_{net}=\int dqEcos\theta$
$={\int }_{-\pi /2}^{\pi /2}\left(\frac{q}{\pi R}\right)Rd\theta \frac{\lambda }{2\pi \epsilon \cdot R}cos\theta$
$=\frac{\lambda q}{2{\pi }^2{\epsilon }_{0}R}{\int }_{-\pi /2}^{\pi /2}cos\theta d\theta =\frac{\lambda q}{2{\pi }^2{\epsilon }_{0}R}[sin\theta {]}_{-\pi /2}^{\pi /2}$
$=\frac{\lambda q}{2{\pi }^2{\epsilon }_{0}R}[1-(-1\left)\right]=\frac{\lambda q}{{\pi }^2{\epsilon }_{0}R}$