Question

The linear charge density of the circumference of a circle of radius $a$ varies as $\lambda = {\lambda _0}\cos \theta$. The total charge on it is …………………
A) $Zero$
B) $Infinite$
C) $\pi a{\lambda _0}$
D) $2\pi a$

Answer 1

To find the solution of this question, first of all we need to integrate the equation of linear charge density. After that we need to put the given value of $\lambda$and solve the equation. The solution of the equation will give the value of total charge on it. Also, we need to find the value of the angle $\theta$by which the radius of the circle is varying.

Complete step by step solution:
We know that linear charge density is given by, $\lambda = \dfrac{{dq}}{{dx}}$ …………………(i)
Where, we know that $\lambda$ is the charge density.
Also, $dq$ is the small amount of charge present on the circumference of the circle,
$dx$ is the circumferential length on which charge is distributed
Equation (i) can be written as,
$dq = \lambda dx$
Now, we need to integrate the above equation.
$\smallint dq = \smallint _0^{2\pi }\lambda dx$
$\Rightarrow q = \smallint _0^{2\pi }{\lambda _0}\cos \theta dx$…………(ii)
We already know that, $\theta = \dfrac{{arc}}{{radius}}$
$\Rightarrow d\theta = \dfrac{{dx}}{a}$
Or, $dx = ad\theta$
Now, putting the value of $dx$in equation (ii), we get,
$q = {\lambda _0}\smallint _0^{2\pi }\cos \theta .ad\theta$
$\Rightarrow q = {\lambda _0}a\smallint _0^{2\pi }\cos \theta .d\theta$
$\Rightarrow q = {\lambda _{_0}}a[\sin \theta ]_0^{2\pi }$
$\Rightarrow q = {\lambda _0}a[\sin 2\pi - \sin 0]$
$\Rightarrow q = {\lambda _0}a[0 - 0]$ $[\because \sin 2\pi = o]$
$\Rightarrow q = {\lambda _0}a[0]$
$\therefore q = 0$

Hence, option (A), i.e. zero is the correct solution for the question.

Note: As, in the given question, we have to find the value of total charge, when linear charge density is given. Linear charge density is the quantity of charge per unit length and is measured in coulombs per meter, at any point on a line charge distribution. Charge can either be positive or negative, since electric charge can be either positive or negative. One should always be clear with the formula of linear charge density. We should not be confused with the surface charge density or volumetric charge density.
Surface charge density, $\sigma = qA$
Volumetric charge density, $\rho = \dfrac{q}{V}$.