Question

The linear charge density of a uniform semicircular wire varies with $\theta$ as shown in the figure as $\lambda ={\lambda }_0\cos \theta$. Find the charge on the wire if $R$ is the radius of the ring.

• A. $3{\lambda }_{0}R$
• B. ${\lambda }_{0}R$
• C. $\frac{2{\lambda }_{0}R}{3}$
• D. $2{\lambda }_{0}R$

Answer 1

The correct option is D $2{\lambda }_{0}R$

Let us choose an element of length $dx$ substending an angle $d\theta$ with the point $\text{O}$.

Charge on the element,
$dq=\lambda dx.....\left(1\right)$
From figure, $dx=Rd\theta$

Substituting the values in $\left(1\right)$,
$dq=\lambda dx={\lambda }_0\cos \theta Rd\theta$

Now total charge on the ring,
$q=\int dq$

$\Rightarrow q=\underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}{\lambda }_0\cos \theta Rd\theta$

$\Rightarrow q={\lambda }_{0}R\underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}\cos \theta d\theta$

$\Rightarrow q={\lambda }_{0}R\left(\sin \theta \right){|}_{-\frac{\pi }{2}}^{+\frac{\pi }{2}}$

$\therefore q=2{\lambda }_{0}R$

Hence, option (d) is the correct answer.