Question

A rod Ab of length L is non-uniformly charged with a liner charge density which depends on distance X from end A of rod as
$\lambda =Cxcoul/m$
Find electric field strength due to this rod a distance r from point B along the axis of rod.

Answer 1

Let AB is the rod having L as its length. Let z is the distance from the middle of the rod. P is the point where electric field is to obtain. Let this distance is x.

The general expression for electric field is

$E\left(P\right)=\frac{1}{4\pi {\epsilon }_0}\underset{line}{\int }\frac{Q}{r^2}$

$\because \lambda =\frac{Q}{L}$

$Q=CxL\left(given\right)$

So, $E\left(P\right)=\frac{1}{4\pi {\epsilon }_0}\underset{line}{\int }\frac{CxL}{r^2}$

Electric field at point P due to E₁ and E₂

$E\left(P\right)=E_1\cos \theta +E_2\cos \theta$

$\because \left|E_1\right|=\left|E_2\right|=E$

$E\left(P\right)=2E\cos \theta$

$E\left(P\right)=\frac{1}{4\pi {\epsilon }_0}\underset{0}{\overset{\frac{l}{2}}{\int }}\frac{Cxdx}{r^2}$

After solving the expression, the final electric field at point P is

$E\left(P\right)=\frac{1}{4\pi {\epsilon }_0}\frac{CL}{z\sqrt{z^2+\frac{L^2}{4}}}$