Question

A hollow sphere of radius 'r' encloses an electric dipole composed of two charges +q and -q. The net flux of electric field through the surface of the sphere due to the enclosed dipole is:

• A. \(\frac{2q}{ε_{0}}\)
• B. \(\frac{2q}{ε_{0}}\)4\(\pi\)r²
• C. infinite
• D. Zero
• E. \(\frac{q}{ε_{0}}\)

Answer 1

The Correct Option is D

A hollow sphere encloses an electric dipole (+q and -q charges).

Gauss's Law

\[ \Phi = \frac{Q_{enc}}{\epsilon_0} \]

Where \( Q_{enc} \) is the net charge enclosed by the surface.

For a dipole: \( Q_{enc} = (+q) + (-q) = 0 \)

Determine Electric Flux \[ \Phi = \frac{0}{\epsilon_0} = 0 \]

Answer 2

1. Recall Gauss's Law:

Gauss's Law states that the net electric flux through any closed surface is equal to the total charge enclosed within the surface divided by the permittivity of free space (\(\epsilon_0\)):

\[\Phi = \frac{Q_{enclosed}}{\epsilon_0}\]

2. Determine the enclosed charge:

The electric dipole consists of two charges: +q and -q. The total charge enclosed by the hollow sphere is the sum of these two charges:

\[Q_{enclosed} = +q + (-q) = 0\]

3. Calculate the net flux:

Since the enclosed charge is zero, the net electric flux through the surface of the sphere is also zero:

\[\Phi = \frac{0}{\epsilon_0} = 0\]