Question

A plane surface of area $200\ \text{cm^2}$ is kept in a uniform electric field of intensity $200\ \text{N/C}$. If the angle between the normal to the surface and the field is $60^\circ$, then the electric flux through the surface is $\left[\cos 60^\circ = \dfrac{1}{2}\right]$

• A. $4\ \text{Nm}^2/\text{C}$
• B. $100\ \text{Nm}^2/\text{C}$
• C. $2\ \text{Nm}^2/\text{C}$
• D. $200\ \text{Nm}^2/\text{C}$

Hint:

Electric flux depends on the component of area perpendicular to the electric field.

Answer 1

The Correct Option is C

Step 1: Write the formula for electric flux.
\[ \Phi = EA\cos\theta \] Step 2: Convert area into SI units.
\[ 200\ \text{cm}^2 = 200 \times 10^{-4}\ \text{m}^2 = 0.02\ \text{m}^2 \] Step 3: Substitute given values.
\[ \Phi = 200 \times 0.02 \times \cos 60^\circ \] Step 4: Calculate electric flux.
\[ \Phi = 200 \times 0.02 \times \dfrac{1}{2} = 2\ \text{Nm}^2/\text{C} \]