Question

A cone lies in a uniform electric field $E$ as shown in figure. The electric flux entering the cone is

• A. $\frac{EhR}{2}$
• B. $E\pi R^2$
• C. $E{h}^2$
• D. $EhR$

Answer 1

The correct option is D $EhR$

Since the electric field is along the horizontal direction, so only that area of the cone whose area vector is parallel to the electric field will contribute to flux.
Plane normal to electric field is a triangle with base length is $2R$ and height is $h$.

Area of triangle $A=\frac{1}{2}\times 2Rh=Rh$
Electric flux entering the cone
$\left(\varphi \right)=EA=ERh$