Question

A long cylinder of radius $R$ is charged with a volume charge density
$\rho =3r^3\text{for}r\le R$
$=0\text{for}$ $r>R$,
where $r$ is the distance measured from the axis of the cylinder, then

• A. The charge contained in $L$ length of the cylinder for $r>R$ is $\frac{6\pi }{5}{R}^{5}L$
• B. The electric field at $r<R$ is $\frac{3r^4}{5{ϵ}_0}$
• C. The electric field at $r>R$ is $\frac{3R^5}{5{ϵ}_{0}r}$
• D. The electric field at $r<R$ is $\frac{2r^4}{5{ϵ}_0}$

Answer 1

Answer: (A), (B), (C)

The correct options are
A The charge contained in $L$ length of the cylinder for $r>R$ is $\frac{6\pi }{5}{R}^{5}L$
B The electric field at $r<R$ is $\frac{3r^4}{5{ϵ}_0}$
C The electric field at $r>R$ is $\frac{3R^5}{5{ϵ}_{0}r}$


$r\le R\rho =\frac{dq}{dv}\Rightarrow dq=\rho \left(dv\right)=\rho \left(2\pi rLdr\right)\Rightarrow dq=\left(3r^3\right)\left(2\pi rLdr\right)\Rightarrow q={\int }_0^{R}6\pi L{r}^{4}dr=\frac{6\pi R^5}{5}L\text{For}r<R,\int Eds=\frac{q_{in}}{{ϵ}_0}=\frac{6\pi L{r}^5}{5{ϵ}_0}\Rightarrow E.2\pi rL=\frac{6\pi L{r}^5}{5{ϵ}_0}\Rightarrow E=\frac{3r^4}{5{ϵ}_0}\text{For}r>R,E.2\pi rL=\frac{6\pi L{R}^5}{5}\Rightarrow E=\frac{3R^5}{5{ϵ}_{0}r}$