# 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?

Gujarat CETGujarat CET 2008Electric Charges And Fields

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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