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:
\(\frac{2q}{ε_{0}}\)
\(\frac{2q}{ε_{0}}\)4\(\pi\)r2
infinite
Zero
\(\frac{q}{ε_{0}}\)
The Correct Option is D
Approach Solution - 1
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 \[ \Phi = \frac{0}{\epsilon_0} = 0 \]
Approach Solution -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:
\[\Phi = \frac{0}{\epsilon_0} = 0\]
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Concepts Used:
Electric Dipole
An electric dipole is a pair of equal and opposite point charges -q and q, separated by a distance of 2a. The direction from q to -q is said to be the direction in space.
p=q×2a
where,
p denotes the electric dipole moment, pointing from the negative charge to the positive charge.
Force Applied on Electric Dipole
