Question

A plane surface area 200 cm² is kept in a uniform electric field of intensity 200 N/C. If the angle between the normal to the surface and the field is 60°, then the electric flux through the surface is

• A. 200 Nm²/C
• B. 4 Nm²/C
• C. 100 Nm²/C
• D. 2 Nm²/C

Answer 1

Answer: (D)

2 Nm²/C

Answer 2

To find the electric flux, we use the formula:

$$\Phi = EA \cos \theta$$

Where:

• $E$ is the electric field intensity.
• $A$ is the area of the surface.
• $\theta$ is the angle between the normal to the surface and the electric field.

First, convert the area from cm² to m²:

$$A = 200 \, \text{cm}^2 = 200 \times 10^{-4} \, \text{m}^2 = 0.02 \, \text{m}^2$$

Now, substitute the given values into the formula:

$$\Phi = 200 \, \text{N/C} \times 0.02 \, \text{m}^2 \times \cos 60^\circ$$

Since $\cos 60^\circ = 0.5$:

$$\Phi = 200 \times 0.02 \times 0.5 = 2 \, \text{Nm}^2/\text{C}$$

Therefore, the electric flux through the surface is $2 \, \text{Nm}^2/\text{C}$.