Question

The number of electric lines of force that radiate outwards from 8.85µC of charges in a vacuum is:

• A. $10^{-6}$
• B. $10^{+6}$
• C. $10^{-12}$
• D. $10^{+12}$

Answer 1

Answer: (B)

10^{+6}

Answer 2

The number of electric lines of force radiating outwards from a charge 'q' in a vacuum is given by the formula:

$N = \frac{q}{\epsilon_0}$

where:

• $q$ is the magnitude of the charge.
• $\epsilon_0$ is the permittivity of free space.

Given:

Charge $q = 8.85 \, \mu C = 8.85 \times 10^{-6} \, C$

Permittivity of free space $\epsilon_0 = 8.85 \times 10^{-12} \, C^2 N^{-1} m^{-2}$

Substitute the given values into the formula:

$N = \frac{8.85 \times 10^{-6} \, C}{8.85 \times 10^{-12} \, C^2 N^{-1} m^{-2}}$

$N = \frac{8.85}{8.85} \times \frac{10^{-6}}{10^{-12}}$

$N = 1 \times 10^{(-6) - (-12)}$

$N = 1 \times 10^{-6 + 12}$

$N = 1 \times 10^6$

Thus, the number of electric lines of force is $10^{+6}$.