Question

A conic surface is placed in a uniform electric field $E$ as shown such that field is perpendicular to the surface on the side AB. The base of the cone is of radius $R$ and height of the cone is $h$. The angle of cone is $\theta$ as shown. Find the magnitude of that flux which enters the cone's curved surface on the left side. Don’t count the outgoing flux.

• A. $ER(hcos\theta +\frac{\pi Rsin\theta }{2})$
• B. $ER(hsin\theta +\frac{\pi Rcos\theta }{2})$
• C. $ER(hcos\theta +\frac{\pi Rsin\theta }{3})$
• D. $ER(hsin\theta +\frac{\pi Rsin\theta }{2})$

Answer 1

The correct option is A

$ER(hcos\theta +\frac{\pi Rsin\theta }{2})$

To find the flux through the curved surface on the left side, One approach could be to try to find the projection of the curved area of the cone perpendicular to electric field but if you think about it, it is not so trivial. Another approach could be to resolve the electric field as shown.

So the projection of the curved surface $\perp$ to the component of electric field $E\cos \theta$ is the triangle with base $2R$ and height $h$
Electric field lines corresponding to the $\cos$ component passing through the curved surface will pass through this triangle and it will be perpendicular to it. So
${\varphi }_1=\left(Ecos\theta \right)$.(area of the triangle)

$Ecos\theta .(\frac{1}{2}\times 2R\times h)$

${\varphi }_1=\left(ERh\cos \theta \right)$

The projection of curved surface perpendicular to $E$$\sin \theta$ is the semi curved disc of the base. So
${\varphi }_2=Esin\theta$ . (area of semi circular disc)

${\varphi }_2=Esin\theta .\left(\frac{\pi R^2}{2}\right)$

$=\frac{\pi E{R}^2}{2}sin\theta$

So Net Flux
$\varphi ={\varphi }_1+{\varphi }_2=ERhcos\theta +\frac{\pi E{R}^2}{2}sin\theta \varphi =ER(hcos\theta +\frac{\pi Rsin\theta }{2})$