Question

A point charge q is placed on vertex of right circular cone. The semi-vertical angle of cone is ${60}^{\circ }$. Find flux of electric field through the base of the cone

• A. $\frac{q}{{ϵ}_0}$
• B. $\frac{q}{2{ϵ}_0}$
• C. $\frac{q}{3{ϵ}_0}$
• D. $\frac{q}{4{ϵ}_0}$

Answer 1

The correct option is D $\frac{q}{4{ϵ}_0}$

Consider Gaussian surface a sphere with its centre at the top and radius, the slant length of cone

Flux through the whole sphere is $\frac{q}{{ϵ}_0}$

Flux through the base of cone $={\varphi }_e=\left(\frac{S}{S_0}\right)\times \frac{q}{{ϵ}_0}$

$S_0=$ Area of the whole sphere

$S=$ Area of sphere below, base of cone

$S_0=4\pi R^2$

$ds=\left(2\pi r\right)Rdr$

integration on $ds=2\pi R.2\sin \theta d\theta$

$\theta ={60}^0$

$S={\int }_0^{\theta }2\pi R^2\sin \theta d\theta =2\pi R^2(1-\cos \theta )$

derives Flux $\varphi =\left(\frac{S}{S_0}\right)\times \frac{q}{{ϵ}_0}=\frac{2\pi R^2(1-\cos \theta )}{4\pi R^2}\left(\frac{q}{{ϵ}_0}\right)$

$\varphi =\frac{q}{{2ϵ}_0}(1-\cos \theta )=\frac{q}{{2ϵ}_0}(1-\frac{1}{2})=\frac{q}{{4ϵ}_0}$

$(\because \theta ={60}^o)$