Question

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?

• A. $2\pi a$
• B. Zero
• C. $\pi a{\lambda }_0$
• D. None of these

Answer 1

Answer: (C)

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
$dq=\lambda \times dl$
$\Rightarrow dq={\lambda }_0{\cos }^2\theta dl$ ..(i)
Now, $d\theta =\frac{dl}{a}\Rightarrow dl=ad\theta$
Then from Eq. (i)
d$q={\lambda }_0{\cos }^2\theta .ad\theta$
$\Rightarrow dq={\lambda }_{0}a{\cos }^2\theta d\theta$ ...(ii)
The total charge on the circle [ on integrating Eq. (ii) with limit 0 to 2 $\pi$]
$q=\underset{0}{\overset{2\pi }{\int }}{\lambda }_{0}a{\cos }^2\theta d\theta$
or $q={\lambda }_{0}a\underset{0}{\overset{2\pi }{\int }}\frac{\cos 2\theta +1}{2}d\theta$
$\left[\because {\cos }^2\theta =\frac{\cos 2\theta +1}{2}\right]$
$\Rightarrow q=\frac{{\lambda }_{0}a}{2}{\left[\frac{\sin 2\theta }{2}+\theta \right]}_0^{2\pi }$
$\Rightarrow q=\frac{{\lambda }_{0}a}{2}\left[\frac{\sin 4\pi }{2}+2\pi -\frac{\sin 0}{2}-0\right]$
$\Rightarrow q=\frac{{\lambda }_{0}a}{2}[0+2\pi -0]$
$\Rightarrow q=\pi a{\lambda }_0$