Question

A rod of length $l$ has a unit positive charge per unit length $\lambda$ and a total charge $Q$. Calculate the electric field at a point P that is located along the long axis of the rod and a distance $a$ from one end.
$k=\frac{1}{4\pi {ϵ}_0}$

• A. $\frac{kQ}{l(l+a)}$
• B. $\frac{kQ}{a(l+a)}$
• C. $\frac{kQ}{a^2}$
• D. $\frac{kQ}{l^2}$

Answer 1

The correct option is B $\frac{kQ}{a(l+a)}$

Take an element of length dx at distance x from point P.

Charge on this element$=dq=\lambda dx$

Field at P due to this element of charge-

$dE=\frac{kdq}{x^2}$

$⟹dE=\frac{k\lambda dx}{x^2}$

$⟹E={\int }_a^{l+a}\frac{k\lambda }{x^2}dx$

$⟹E=k\lambda {\left[\frac{-1}{x}\right]}_a^{l+a}$

$⟹E=k\lambda (\frac{1}{a}-\frac{1}{l+a})$

$⟹E=\frac{k\lambda l}{a(a+l)}$

Total charge on the rod$=Q=\lambda l$

Hence, $E=\frac{kQ}{a(a+l)}$

Answer-(B)