Q199 / 200 · MHT CET - Medical 2026 (25 April Shift 1) · Physics
A closed surface encloses $n$ electric dipoles. The net electric flux through the closed surface is
($\epsilon_0$ = permittivity of free space, q = charge)
($\epsilon_0$ = permittivity of free space, q = charge)
- A$\dfrac{nq}{\epsilon_0}$
- B$n\epsilon_0 q$
- C$\dfrac{q}{\epsilon_0}$
- Correct answerD$0$
Solution
An electric dipole consists of two equal and opposite charges, $+q$ and $-q$.
The net charge of a single electric dipole is $+q - q = 0$.
For $n$ electric dipoles enclosed in a closed surface, the total enclosed charge is $Q_{enclosed} = n \times 0 = 0$.
According to Gauss's law, the net electric flux through a closed surface is given by $\Phi = \dfrac{Q_{enclosed}}{\epsilon_0}$.
Substituting $Q_{enclosed} = 0$, we get $\Phi = 0$.
Answer: $0$
The net charge of a single electric dipole is $+q - q = 0$.
For $n$ electric dipoles enclosed in a closed surface, the total enclosed charge is $Q_{enclosed} = n \times 0 = 0$.
According to Gauss's law, the net electric flux through a closed surface is given by $\Phi = \dfrac{Q_{enclosed}}{\epsilon_0}$.
Substituting $Q_{enclosed} = 0$, we get $\Phi = 0$.
Answer: $0$
Hence, the correct answer is option: 'D'.