Question

Axis of a hollow cone shown in figure is vertical. It’s base radius is ‘R’. It is kept in a uniform electric field E parallel to its axis.

The magnitude of flux through curved part of the cone?

• A. $\pi R^{2}E$
• B. $\frac{\pi R^{2}E}{2}$
• C. $zero$
• D. $Non-zero$

Answer 1

Answer: (A), (D)

$Non-zero$

As the field is parallel to the axis, clearly

flux $\ne 0\Rightarrow \left(d\right)$

The field is constant. The flux through the curved part is equal to the flux through the circular base. Why?

Mathematically if you evaluate $\overrightarrow{E}.\overrightarrow{dA}$ i.e., the projection of curved part of the cone onto its opposite plane you will obtain the same result.

By the help of Electric field lines it is much simpler to infer.

Hence, ${\varphi }_{curevedpart}=E\times \pi R^2$