Question

As shown in the figure, a surface encloses an electric dipole with charges $\pm6 \times 10^{-6} \, {C}$. The total electric flux through the closed surface is:

 

• A. $+12 \times 10^{-6} \, {Nm}^2{C}^{-1}$
• B. $-12 \times 10^{-6} \, {Nm}^2{C}^{-1}$
• C. Zero
• D. $+6 \times 10^{-6} \, {Nm}^2{C}^{-1}$

Hint:

Remember that for any closed surface enclosing a dipole, the net electric flux is always zero, regardless of the dipole's orientation within the surface. This is a direct result of Gauss's Law.

Answer 1

The Correct Option is C

Step 1: Apply Gauss's Law for a closed surface. Gauss's Law states that the total electric flux through a closed surface is proportional to the charge enclosed within that surface. For a dipole with charges $\pm6 \times 10^{-6} \, {C}$, the net charge enclosed is zero: $$Q_{{enc}} = +6 \times 10^{-6} \, {C} + (-6 \times 10^{-6} \, {C}) = 0 \, {C}$$ Step 2: Calculate the electric flux. Since the net enclosed charge $Q_{{enc}}$ is zero, the electric flux $\Phi$ through the surface is: $$\Phi = \frac{Q_{{enc}}}{\epsilon_0} = \frac{0 \, {C}}{\epsilon_0} = 0 \, {Nm}^2{C}^{-1}$$