Question

In fig., a cone lies in a uniform electric field E. Determine the electric flux entering the cone.

• A. Flux $\varphi =EA=2ERh$
• B. Flux $\varphi =EA=7ERh$
• C. Flux $\varphi =EA=9ERh$
• D. Flux $\varphi =EA=ERh$

Answer 1

The correct option is D Flux $\varphi =EA=ERh$

$\text{Step 1: Representation of cross-sectional area}$

Electric flux is proportional to number of field lines crossing the surface.

Here, Number of field lines entering the curved surface equals the number of lines crossing the corresponding cross section area shown in the figure.

Hence Flux entering the cone equals the flux crossing the cross section area.

$\text{Step 2: Calculation of electric flux}$

We know that, Electric Flux $\varphi =EA\cos \theta$

Where $\theta$ is angle between E and A

Here $\theta =0^o$ as Electric field is crossing the Area perpendicularly i.e. $\overrightarrow{E}$ is perpendicular to the $\overrightarrow{A}$, as Area vector is perpendicular to the plane of the Area.

And Area of triangle, $A=\frac{1}{2}\times Base\times Height$

$=\frac{1}{2}\times 2R\times h=Rh$

So, Flux $\varphi =E\left(Rh\right)\cos 0^o$

$\Rightarrow \varphi =ERh$

Hence option 'D' is correct