A circular plate sheet of radius $10 \mathrm{~cm}$ is placed in a uniform electric field of $2 \sqrt{3} \times 10^5 \mathrm{NC}^{-1}$, making an angle of $60^{\circ}$ with the field. Then, find the electric flux through the sheet.

• A

  $1.36 \times 10^2 \mathrm{Nm}^2 \mathrm{C}^{-1}$
• Correct answer
  B

  $9.42 \times 10^3 \mathrm{Nm}^2 \mathrm{C}^{-1}$
• C

  $0.515 \times 10^2 \mathrm{Nm}^2 \mathrm{C}^{-1}$
• D

  $0.515 \times 10^4 \mathrm{Nm}^2 \mathrm{C}^{-1}$

Solution

Given that, radius of plate, $$ \begin{aligned} R & =10 \mathrm{~cm} \\ & =10 \times 10^{-2} \mathrm{~m} \end{aligned} $$ Uniform electric field, $E=2 \sqrt{3} \times 10^5 \mathrm{NC}^{-1}$ Since, the angle between plate and electric field is $60^{\circ}$.
Then, angle between normal to the plate and electric field, $\theta=90^{\circ}-60^{\circ}=30^{\circ}$. By using expression of electric flux, $$ \begin{aligned} \phi & =E A \cos \theta=2 \sqrt{3} \times 10^5 \times \pi\left(10 \times 10^{-2}\right)^2 \times \cos 30^{\circ} \\ & =9.42 \times 10^3 \mathrm{~N}-\mathrm{m}^2 \mathrm{C}^{-1} \end{aligned} $$

Hence, the correct answer is option: 'B'.