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 calculate the electric flux, we use the formula:

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

Where:

• $E$ is the electric field intensity (200 N/C).
• $A$ is the area of the surface in m². First, convert the area from cm² to m²: $$200 \text{ cm}^2 = 200 \times 10^{-4} \text{ m}^2 = 0.02 \text{ m}^2$$
• $\theta$ is the angle between the normal to the surface and the electric field (60°).

Now, plug in the values:

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

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

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

Thus, the electric flux through the surface is 2 Nm²/C.