A long cylinder contains charge distributed uniformly having volume charge density ρ.
What would be the electric field inside the cylinder if the volume charge density of the cylinder varies according to the relation ρ = kx2?
Answer & explanation
Correct answer: option 3
$ Q_{in} = \int_0^x{\rho 2\pi lxdx} = \int_0^x{kx^2 2\pi lxdx} = \frac{k\pi l x^4}{2} $
$ \text{Using gauss law } E.2\pi xl = \frac{q_{in}}{\epsilon_0} = \frac{k\pi l x^4}{2\epsilon_0} $
$\Rightarrow E = \frac{kx^3}{4\epsilon_0}$