Three infinite plane sheets which have uniform positive surface charge densities $\sigma$, $\sigma$ and $2\sigma$, are arranged parallel to each other with a separation of $d$ as shown in the figure. A spherical Gaussian surface $S$ of radius $d/2$ has its center on the middle sheet. Which of the following statements regarding the electric flux $\Phi_L$ through the left hemisphere and the electric flux $\Phi_R$ through the right hemisphere of the Gaussian surface is correct?
Show Hint
Using superposition to find the net electric field in each region is a quick way to solve flux problems.
If the field is zero in a region, the flux through the corresponding bounding surface in that region is also zero.
Step 1: Understanding the Question:
We have three parallel infinite charged sheets of surface charge densities $\sigma$ (left), $\sigma$ (middle), and $2\sigma$ (right).
A spherical Gaussian surface of radius $d/2$ is centered on the middle sheet.
We need to compare the flux through the left and right hemispheres of this sphere. Step 2: Key Formula or Approach:
The electric field due to an infinite plane sheet of charge density $\sigma$ is:
\[ E = \frac{\sigma}{2\varepsilon_0} \]
directed away from the positive sheet.
Using the superposition principle, we find the net electric field in the regions inside the sphere on both sides of the middle sheet. Step 3: Detailed Explanation:
• Let us define the positions of the sheets on the $x$-axis:
- Left sheet: $x = -d$, with $\sigma_1 = \sigma$
- Middle sheet: $x = 0$, with $\sigma_2 = \sigma$
- Right sheet: $x = d$, with $\sigma_3 = 2\sigma$
• The Gaussian sphere of radius $d/2$ is centered at $x = 0$.
- The left hemisphere lies in the region $-d/2 < x < 0$.
- The right hemisphere lies in the region $0 < x < d/2$.
• Let us calculate the electric field in the left region ($-d/2 < x < 0$):
- Field due to left sheet: $\mathbf{E}_1 = +\frac{\sigma}{2\varepsilon_0} \hat{i}$ (pointing right)
- Field due to middle sheet: $\mathbf{E}_2 = -\frac{\sigma}{2\varepsilon_0} \hat{i}$ (pointing left)
- Field due to right sheet: $\mathbf{E}_3 = -\frac{2\sigma}{2\varepsilon_0} \hat{i}$ (pointing left)
- Net field:
\[ \mathbf{E}_L = \left( \frac{\sigma}{2\varepsilon_0} - \frac{\sigma}{2\varepsilon_0} - \frac{2\sigma}{2\varepsilon_0} \right) \hat{i} = -\frac{\sigma}{\varepsilon_0} \hat{i} \]
Thus, there is a non-zero electric field pointing to the left in the left hemisphere.
• Let us calculate the electric field in the right region ($0 < x < d/2$):
- Field due to left sheet: $\mathbf{E}_1 = +\frac{\sigma}{2\varepsilon_0} \hat{i}$ (pointing right)
- Field due to middle sheet: $\mathbf{E}_2 = +\frac{\sigma}{2\varepsilon_0} \hat{i}$ (pointing right)
- Field due to right sheet: $\mathbf{E}_3 = -\frac{2\sigma}{2\varepsilon_0} \hat{i}$ (pointing left)
- Net field:
\[ \mathbf{E}_R = \left( \frac{\sigma}{2\varepsilon_0} + \frac{\sigma}{2\varepsilon_0} - \frac{2\sigma}{2\varepsilon_0} \right) \hat{i} = \mathbf{0} \]
Thus, the electric field is exactly zero in the region of the right hemisphere.
• Since $\mathbf{E}_R = \mathbf{0}$, the flux through the right hemisphere is $\Phi_R = 0$.
• Since $\mathbf{E}_L \neq \mathbf{0}$ and points outward through the curved surface, the flux through the left hemisphere is positive ($\Phi_L > 0$).
• Therefore, $\Phi_L > \Phi_R$. Step 4: Final Answer:
The electric flux through the left hemisphere is greater than that through the right hemisphere ($\Phi_L > \Phi_R$).
