Question

Point charge $q_0$ is placed inside a cone of base radius 'R', x distance below centre of the top surface as shown in figure.Find electric flux related to curved surface of the cone:-

• A. $\frac{q}{2{\epsilon }_0}$
• B. $\frac{q}{{\epsilon }_0}-\frac{q}{2{\epsilon }_0}\left(1-\frac{x}{\sqrt{R^2+x^2}}\right)$
• C. $\frac{q}{{\epsilon }_0}\left(1-\frac{x}{\sqrt{R^2+x^2}}\right)$
• D. $\frac{q}{{\epsilon }_0}\times \left(\frac{x}{\sqrt{R^2+x^2}}\right)$

Answer 1

The correct option is A $\frac{q}{2{\epsilon }_0}$

Point charge $=q$

radius $=R$

distance $=x$

Consider a Gaussian surface, a sphere with its cetnre at the top and radius the start length of the cone. The flux through the whole sphere is $\frac{q}{{ϵ}_0}$. Therefore, the flux through the base of the cone is.

${\varphi }_e=\left(\frac{s}{s_0}\right)\left(\frac{q}{{ϵ}_0}\right)$

$s_0=$ area of whole sphere

$s=$ area of sphere below the base of cone

Because all those field lines which pass through the base of the cone will pass through the cap of sphere.

Let, $R=$ radius of Gaussian sphere

$s_0=$ area of whole sphere $=4\pi R^2$

$s=$ area of sphere below the base of the cone which can be found by integration.

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

$ds=2\pi R^2\sin \alpha d\alpha$

$s={\int }_0^{\theta }ds={\int }_0^{\theta }2\pi R^2\sin \alpha d\alpha =2\pi R^2$

The desired flux

${\varphi }_e=\frac{s}{s_0}.\frac{q}{{ϵ}_0}$

$=\frac{2\pi R^2}{4\pi R^2}.\frac{q}{{ϵ}_0}$

$=\frac{q}{{2ϵ}_0}$