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: \[ \Phi = \frac{Q_{{enc}}}{\epsilon_0} = \frac{0 \, {C}}{\epsilon_0} = 0 \, {Nm}^2{C}^{-1} \]