The total electric flux through a closed spherical surface of radius ‘r’ enclosing an electric dipole of dipole moment 2aq is (Give ε 0 = permittivity of free space)

Question

The total electric flux through a closed spherical surface of radius &lsquo;r&rsquo; enclosing an electric dipole of dipole moment 2aq is (Give &epsilon; 0 = permittivity of free space)

cdquestions Admin · Accepted Answer

The given question is about the electric flux through a closed spherical surface enclosing an electric dipole. Step 1: Apply Gauss's Law The electric flux $ \Phi_E $ is given by Gauss's law: $$ \Phi_E = \frac{q_{\text{enclosed}}}{\epsilon_0} $$ Step 2: Understanding the Dipole A dipole consists of two equal and opposite charges, so the total charge enclosed by the spherical surface is zero. Thus, the flux is zero: $$ \Phi_E = 0 $$ The correct answer is (A) Zero .

Answer

Answer

$\frac{q}{\epsilon_0}$

Answer

$\frac{2q}{\epsilon_0}$

Answer

$\frac{8\pi r^2q}{\epsilon_0}$

LIST I		LIST II
A	Gauss's Law in Electrostatics	I	\(\oint \vec{E} \cdot d \vec{l}=-\frac{d \phi_B}{d t}\)
B	Faraday's Law	II	\(\oint \vec{B} \cdot d \vec{A}=0\)
C	Gauss's Law in Magnetism	III	\(\oint \vec{B} \cdot d \vec{l}=\mu_0 i_c+\mu_0 \in_0 \frac{d \phi_E}{d t}\)
D	Ampere-Maxwell Law	IV	\(\oint \vec{E} \cdot d \vec{s}=\frac{q}{\epsilon_0}\)

The total electric flux through a closed spherical surface of radius ‘r’ enclosing an electric dipole of dipole moment 2aq is (Give ε₀ = permittivity of free space)

The Correct Option is A

Solution and Explanation

Step 1: Apply Gauss's Law

Step 2: Understanding the Dipole

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The total electric flux through a closed spherical surface of radius ‘r’ enclosing an electric dipole of dipole moment 2aq is (Give ε0 = permittivity of free space)