Q. A circle of radius a has charge density given by $ \lambda = \lambda_0 \ cos^2 \ \theta$ on its circumference. What will be the total charge on the circle?

Solution:

Charge density
Let PQ be an element of length dl on the circumference of the circle which makes an angle d$\theta$ at the centre of the circle.
Then the charge on the element is
dq = $ \lambda \times dl$
$\Rightarrow dq = \lambda_0 \ cos^2 \ \theta.dl$ $ \hspace20mm$ ..(i)
Now, $ d \theta = \frac{ dl}{ a} \Rightarrow dl = ad \ \theta $
Then from Eq. (i)
dq = $ \lambda_0 \ cos^2 \ \theta . ad \ \theta $
$\Rightarrow dq = \lambda_0 a \ \ cos^2 \ \theta.d \theta$ $ \hspace20mm$ ..(ii)
The total charge on the circle [ on integrating Eq. (ii) with limit 0 to 2 $\pi$]
q = $ \int \limits_0^{ 2 \pi} \lambda_0 a \ cos^2 \ \theta d \ \theta$
or $ q = \lambda_0 a \int \limits_0^{ 2 \pi} \frac{ cos \ 2 \theta + 1}{ 2} d \theta $
$\hspace20mm$ $ \bigg [ \because cos^2 \ \theta = \frac{ cos \ 2 \theta + 1}{ 2} \bigg ] $
$\Rightarrow q = \frac{ \lambda_0 a }{ 2} \bigg[ \frac{ sin \ 2 \theta }{ 2} + \theta \bigg ] _0^{ 2 \pi} $
$\Rightarrow q = \frac{ \lambda_0 a }{ 2} \bigg[ \frac{ sin \ 4 \pi}{ 2} + 2 \pi - \frac{ sin \ 0 }{ 2} - 0 \bigg ] $
$\Rightarrow q = \frac{ \lambda_0 a }{ 2} [ 0 + 2\pi - 0 ] $
$\Rightarrow q = \pi a \lambda_0$

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