Question

The axis of a hollow cone is vertical.
Its base radius is R. It is kept in a
uniform electric field E
parallel to its
axis. Magnitude of flux through curved
surface MNQP of cone is

Answer 1

Given that,

Base of radius = R

Electric field = E

The base of the cone is of radius R, and the height of the cone is h. The angle of the cone is θ.

We need to calculate the magnitude of flux through curved surface MNQP of cone

Using formula of flux

$\phi=EA\cos\theta$

Flux entering the cone from side MN will ultimately also pass through areas A₁ and A₂ .

So,

$\phi=EA_{1}\cos\theta+EA_{2}\cos(90-\theta)$

Put the value into the formula

$\phi=E(\dfrac{1}{2}\times2Rh\cos\theta+\dfrac{\pi}{2}R^2\sin\theta)$

$\phi=ER(h\cos\theta+\pi\times\dfrac{R}{2}\sin\theta)$

Hence, The magnitude of flux through curved surface MNQP of cone is $ER(h\cos\theta+\pi\times\dfrac{R}{2}\sin\theta)$