Six infinitely large and thin non-conducting sheets are fixed in configurations I and II. As shown in the figure, the sheets carry uniform surface charge densities which are indicated in terms of $\sigma_0$. The separation between any two consecutive sheets is $1 \, \mu m$. The various regions between the sheets are denoted as 1, 2, 3, 4 and 5. If $\sigma_0 = 9 \, \mu C/m^2$, then which of the following statements is/are correct? (Take permittivity of free space $\epsilon_0 = 9 \times 10^{-12} \, \text{F/m}$)
Show Hint
- In region 4 of the configuration I, the magnitude of the electric field is zero.
- In region 3 of the configuration II, the magnitude of the electric field is \(\frac{\sigma_0}{\epsilon_0}\).
- Potential difference between the first and the last sheets of the configuration I is 5 V.
- Potential difference between the first and the last sheets of the configuration II is zero.
The Correct Option is A, B, C
Solution and Explanation
We have two configurations of six charged sheets with given surface charge densities. The sheets create electric fields in the regions between them. We need to find the electric field magnitudes in specific regions and the potential differences between the first and last sheets.
Step 2: Electric field due to a charged sheet
The electric field due to an infinite sheet with surface charge density \(\sigma\) is: \[ E = \frac{\sigma}{2\epsilon_0} \] pointing away from the positively charged sheet and toward the negatively charged sheet.
Step 3: Calculate electric fields in configuration I
Charges on sheets are \(+\sigma_0, -\sigma_0, +\sigma_0, -\sigma_0, +\sigma_0, -\sigma_0\).
- Region 4 lies between the 4th and 5th sheets which have charges \(-\sigma_0\) and \(+\sigma_0\).
- The electric fields from adjacent sheets cancel because they are equal and opposite in direction in region 4.
- Thus, magnitude of electric field in region 4 is zero.
Step 4: Calculate electric fields in configuration II
Charges on sheets are \(\frac{+\sigma_0}{2}, -\sigma_0, +\sigma_0, -\sigma_0, +\sigma_0, \frac{-\sigma_0}{2}\).
- Consider region 3 (between sheets 3 and 4 with charges \(+\sigma_0\) and \(-\sigma_0\)).
- Electric fields add up due to asymmetric charges.
- Total electric field magnitude in region 3 is \(\frac{\sigma_0}{\epsilon_0}\).
- Distance between first and last sheets: \(5 \times 1 \, \mu m = 5 \times 10^{-6} \, m\).
- Electric field in each region (except region 4) is \(\frac{\sigma_0}{\epsilon_0}\) or 0 in region 4.
- Summing potential drops across regions: \[ V = \sum E \cdot d = 5 \times \frac{\sigma_0}{\epsilon_0} \times 10^{-6} \] Given \(\sigma_0 = 9 \times 10^{-6} \, C/m^2\), \(\epsilon_0 = 9 \times 10^{-12} \, F/m\), \[ V = 5 \times \frac{9 \times 10^{-6}}{9 \times 10^{-12}} \times 10^{-6} = 5 \, V \]
- The net electric field across the entire configuration II sums to zero due to the fractional charges on the end sheets balancing the others.
- Thus, the potential difference between the first and last sheets is zero.
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