Question 838: Electrostats & Capacitance

Question

A charge $$17.7 	imes 10^{-4} \mathrm{C}$$ is distributed uniformly over a large sheet of area $$200 \mathrm{~m}^2$$. The electric field intensity at a distance $$20 \mathrm{~cm}$$ from it in air will be $$\left[\varepsilon_0=8.85 	imes 10^{-12} \mathrm{C}^2 / \mathrm{Nm}^2ight]$$

Accepted Answer

The surface charge density is given by,
$$\sigma=\frac{\mathrm{q}}{\mathrm{A}}=\frac{17.7 	imes 10^{-4}}{200}=8.85 	imes 10^{-6} \mathrm{C} / \mathrm{m}^2$$
The electric field intensity at a distance of $$20 \mathrm{~cm}$$ in air is,
$$\mathrm{E}=\frac{\sigma}{2 \varepsilon_0}=\frac{8.85 	imes 10^{-6}}{2 	imes 8.85 	imes 10^{-12}}=5 	imes 10^5 \mathrm{~N} / \mathrm{C}$$
Hence, option (A).

Answer

$$5 	imes 10^5 \mathrm{~N} / \mathrm{C}$$

Answer

$$6 	imes 10^5 \mathrm{~N} / \mathrm{C}$$

Answer

$$7 	imes 10^5 \mathrm{~N} / \mathrm{C}$$

Answer

$$8 	imes 10^5 \mathrm{~N} / \mathrm{C}$$

Electrostats & Capacitance

A charge $17.7 \times 10^{-4} \mathrm{C} is distributed uniformly over a large sheet of area 200 \mathrm{~m}^2. The electric field intensity at a distance 20 \mathrm{~cm} from it in air will be \left[\varepsilon_0=8.85 \times 10^{-12} \mathrm{C}^2 / \mathrm{Nm}^2\right]

Choose the correct answer: