Question

If a rectangular area is rotated in a uniform electric field from the position where the maximum electric flux goes through it to an orientation where only half the flux goes through it, what has
been the angle of rotation?

Answer 1

Concepts and reason

The main concepts required to solve this problem are electric field, area and electric flux.

Initially, write the concept and equation for the electric flux. Use this equation and then calculate the angle of rotation.

Fundamentals

Electric flux can be defined as the dot product of the electric field and the areal vector. Electric flux can also be defined as the number of electric field lines passing through the unit surface area.

The equation for the electric flux is,

${\phi _{{\rm{max}}}} = EA{\rm{cos}}\theta$

Here, E is the electric field, A is the areal vector and $\theta$ is the angle between the electric field direction and the areal vector.

The expression for the electric flux is,

$\phi = EA{\rm{cos}}\theta$ …… (1)

If the electric field direction is parallel to the surface area, then the flux will be maximum.

The angle between the electric field and the area when they are parallel is,

$\theta = {0^{\rm{o}}}$

Substitute ${0^{\rm{o}}}$ for $\theta$ in above equation (1).

$\begin{array}{c}\\{\phi _{{\rm{max}}}} = EA{\rm{cos}}{0^{\rm{o}}}\\\\ = EA\left( 1 \right)\\\\ = EA\\\end{array}$

Let the angle of rotation be $\theta$ .

When the flux is oriented at some angle then electric flux is equal to the half of the maximum electric flux.

After the orientation, the flux is,

$\phi = EA{\rm{cos}}\theta$

According to the condition that given in the problem, the maximum flux is half to the flux after the orientation, that is,

$\phi = \frac{{{\phi _{{\rm{max}}}}}}{2}$

Substitute $EA{\rm{cos}}\theta$ for $\phi$ , and $EA$ for ${\phi _{{\rm{max}}}}$ in above equation.

$\begin{array}{c}\\EA{\rm{cos}}\theta = \frac{1}{2}EA\\\\{\rm{cos}}\theta = \frac{1}{2}\\\\\theta = {\cos ^{ - 1}}\left( {\frac{1}{2}} \right)\\\\ = {60^{\rm{o}}}\\\end{array}$

Ans:

The angle of rotation of the electric flux is ${60^{\rm{o}}}$ .