Question

A square frame of edge $5\text{cm}$ is placed with its positive normal making an angle of ${60}^{\circ }$ with a uniform electric field of $200\text{N/C}$. Find the electric flux through the surface bounded by the frame.

• A. $0.25{\text{N-m}}^2/\text{C}$
• B. $0.45{\text{N-m}}^2/\text{C}$
• C. $0.15{\text{N-m}}^2/\text{C}$
• D. $0.35{\text{N-m}}^2/\text{C}$

Answer 1

The correct option is A $0.25{\text{N-m}}^2/\text{C}$

Given:
Edge of square frame, $a=5\text{cm}$
So, area of square frame, $A=a^2=25{\text{cm}}^2=25\times {10}^{-4}{\text{m}}^2$

Angle made by positive normal of area with electric field, $\theta ={60}^{\circ }$

Magnitude of electric field, $E=200\text{N/C}$


We know that electric flux $\varphi$ is given by
$\varphi =\overrightarrow{E}.\overrightarrow{A}=EA\cos \theta$

$\Rightarrow \varphi =200\times 25\times {10}^{-4}\times \cos {60}^{\circ }$

$\Rightarrow \varphi =0.25{\text{N-m}}^2/\text{C}$