Question

A cone made of insulating material has a total charge $Q$ spread uniformly over its sloping surface having a slant length $L$ . The energy required to take a test charge $q$ from infinity to apex $A$ of the cone will be

• A. $\frac{Qq}{2\pi {\epsilon }_{0}L}$
• B. $\frac{Qq}{4\pi {\epsilon }_{0}L}$
• C. $-\frac{Qq}{3\pi {\epsilon }_{0}L}$
• D. $\frac{Qq}{8\pi {\epsilon }_{0}L}$

Answer 1

Answer: (A)

$\sigma =\frac{Q}{\pi RL}$
$L$ = Slant height
$dQ=\sigma \times \left(2\pi r\right)dx$
= Charge on the elementary ring
$dQ=\frac{Q}{\pi RL}\times 2\pi rdx=\frac{Q}{LR}\left(2rdx\right)$ ... (1)
By similarity of triangles :
$\frac{R}{L}=\frac{r}{x}\Rightarrow r=\frac{R}{L}x$ ... (2)
By (1) and (2) we get
$dQ=\frac{Q}{LR}\times 2\frac{R}{L}xdx=\frac{2Q}{L^2}xdx$
$\therefore$ the electric potential at A due to ring is
$dV=\frac{kdQ}{x}=\frac{k}{x}\frac{\left(2Q\right)}{L^2}xdx$
$\therefore$ Total potential at A is
$V=\int dv=\frac{2kQ}{L^2}{\int }_0^{L}dx=\frac{2kQ}{L^2}\times L$
$V=\frac{2kQ}{L}$
$\therefore$ Work done to shift a charge q from infinity to point $A$ is = $qV$
$W=\frac{2kqQ}{L}=2\times \frac{1}{4\pi {ϵ}_0}\frac{qQ}{L}=\frac{qQ}{2\pi {ϵ}_{0}L}$