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 Concept:
The flux through a surface depends on the magnitude and direction of the net electric field vector integrated over that surface. Step 2: Detailed Explanation:
1. Field from one sheet: \( E = \frac{\sigma}{2\epsilon_{0}} \).
2. Left region (between left and middle sheets):
$\bullet$ From Left sheet (\( \sigma \)): Field is \( E_{0} \) to the right.
$\bullet$ From Middle sheet (\( \sigma \)): Field is \( E_{0} \) to the left.
$\bullet$ From Right sheet (\( 2\sigma \)): Field is \( 2E_{0} \) to the left.
Net field \( E_{L} = E_{0} - E_{0} - 2E_{0} = -2E_{0} \) (towards left).
3. Right region (between middle and right sheets):
$\bullet$ From Left sheet: \( E_{0} \) to right.
$\bullet$ From Middle sheet: \( E_{0} \) to right.
$\bullet$ From Right sheet: \( 2E_{0} \) to left.
Net field \( E_{R} = E_{0} + E_{0} - 2E_{0} = 0 \).
4. Flux comparison:
$\Phi_{R}$ is 0 because the field in that region is zero. $\Phi_{L}$ is non-zero (and positive as it points out of the sphere). Therefore \( \Phi_{L} > \Phi_{R} \). Step 3: Final Answer:
The flux through the left hemisphere is greater.
