Question

The volume charge density of a sphere of radius 6 m is 2μC cm^-3. The number of lines of force per unit surface area coming out from the surface of the sphere is _______ × 10¹⁰ NC^-1.

Answer 1

Correct Answer: 45

 To solve the problem, we need to determine the number of lines of force per unit surface area coming out from the surface of the sphere. This can be computed using Gauss's law in electrostatics, which relates the electric flux through a closed surface to the charge enclosed by the surface. Here are the steps:

Step 1: Calculate the total charge within the sphere.
Given, the radius of the sphere \( R = 6 \, \text{m} = 600 \, \text{cm} \), volume charge density \( \rho = 2 \, \mu\text{C} \, \text{cm}^{-3} = 2 \times 10^{-6} \, \text{C} \, \text{cm}^{-3} \).
The volume of the sphere \( V = \frac{4}{3} \pi R^3 = \frac{4}{3} \pi (600)^3 \, \text{cm}^3 \).
Total charge \( Q = \rho \times V = 2 \times 10^{-6} \times \frac{4}{3} \pi (600)^3 \) (Coulombs).

Step 2: Calculate the electric field at the surface of the sphere.
By Gauss's law, the electric flux \( \Phi = E \cdot 4 \pi R^2 = \frac{Q}{\varepsilon_0} \), where \( \varepsilon_0 \) is the permittivity of free space (\( \varepsilon_0 = 8.854 \times 10^{-12} \, \text{C}^2/\text{N}\cdot\text{m}^2 \)).
Hence, the electric field \( E = \frac{Q}{4 \pi R^2 \varepsilon_0} \).

Step 3: Find the number of lines of force per unit area.
The number of lines of force per unit surface area \( = E/\varepsilon_0 = \frac{Q}{4 \pi R^2 \varepsilon_0^2} \).
Calculate this as:
First find \( Q \) by substituting the known values:
\( Q = 2 \times 10^{-6} \times \frac{4}{3} \pi (600)^3 \, = 2 \times 10^{-6} \times \frac{4}{3} \times 3.14159 \times 2.16 \times 10^8 \, \approx 1.8096 \times 10^3 \, \text{C} \).
Substitute \( Q \) and other constants in \( \frac{Q}{4 \pi R^2 \varepsilon_0^2} \):
\( \text{Value} = \frac{1.8096 \times 10^3}{4 \pi (600)^2 (8.854 \times 10^{-12})^2} \approx 4.5 \times 10^{11} \, \text{NC}^{-1} \).
The given result is indeed \( 45 \times 10^{10} \, \text{NC}^{-1} \), falling within the expected range \( 45,45 \).
The number of lines of force per unit surface area is \( 45 \times 10^{10} \, \text{NC}^{-1} \), concluding our solution.