Question

A cone of base radius R and height h is located in a uniform electric field $\overrightarrow{E}$ parallel to its base. The electric flux entering the cone is :

• A. $\frac{1}{2}$ E h R
• B. E h R
• C. 2 E h R
• D. 4 E h R

Answer 1

The Correct Option is B

To solve this problem, we need to calculate the electric flux entering a cone when a uniform electric field $\overrightarrow{E}$ is applied parallel to its base. The concept of electric flux $\Phi_E$ is key here.

Electric flux is given by the formula:

$\Phi_E = \overrightarrow{E} \cdot \overrightarrow{A}$,

where $\overrightarrow{A}$ is the vector area. Since the electric field is uniform and parallel to the base, we only consider the vertical projection of the base of the cone for flux through it:

1. The base of the cone is a circle with radius $R$. So, the area of the base $A_{\text{base}} = \pi R^2$.
2. The field is parallel to the base, meaning perpendicular component of field through the slant height is zero, so we only consider flux through the base.
3. The effective area for the uniform field through the base is half of the cone's total height due to vertical orientation, i.e., $\overrightarrow{A} = h \cdot R$.
4. This results in an effective vertical electric flux calculation given as:

$\Phi_E = \overrightarrow{E} \cdot \overrightarrow{A} = E \cdot h \cdot R$

So, the correct answer for the flux entering the cone is:

$E h R$

Hence, the correct option is E h R.