Electrostatics Question 724
Question: Find the force experienced by a semicircular rod having a charge q as shown in Fig. 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.
Options:
A) $ \frac{\lambda q}{2{{\pi }^{2}}{\varepsilon _{0}}R} $
B) $ \frac{\lambda q}{{{\pi }^{2}}{\varepsilon _{0}}R} $
C) $ \frac{\lambda q}{4{{\pi }^{2}}{\varepsilon _{0}}R} $
D) $ \frac{\lambda q}{4\pi {\varepsilon _{0}}R} $
Show Answer
Answer:
Correct Answer: B
Solution:
[b] $ F _{net}=\int _{{}}^{{}}{dqE\cos \theta } $
$ =\int\limits _{-\pi /2}^{\pi /2}{( \frac{q}{\pi R} )Rd\theta \frac{\lambda }{2\pi \varepsilon R}\cos \theta } $
$ =\frac{\lambda q}{2{{\pi }^{2}}{\varepsilon _{0}}R}\int\limits _{-\pi /2}^{\pi /2}{\cos \theta d\theta =\frac{\lambda q}{2{{\pi }^{2}}{\varepsilon _{0}}R}[ \sin \theta ] _{-\pi /2}^{\pi /2}} $
$ =\frac{\lambda q}{2{{\pi }^{2}}{\varepsilon _{0}}R}[1-(-1)]=\frac{\lambda q}{{{\pi }^{2}}{\varepsilon _{0}}R} $
BETA
