Question

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 σ₀. The separation between any two consecutive sheets is 1 µm. The various regions between the sheets are denoted as 1, 2, 3, 4 and 5. If σ₀ = 9 µC/m², then which of the following statements is/are correct?

(Take permittivity of free space ε₀ = 9 × 10⁻¹² F/m)

• A. In region 4 of the configuration I, the magnitude of the electric field is zero.
• B. In region 3 of the configuration II, the magnitude of the electric field is $\frac{σ₀}{ε₀}$.
• C. Potential difference between the first and the last sheets of the configuration I is 5 V.
• D. Potential difference between the first and the last sheets of the configuration II is zero.

Answer 1

Answer: (B)

B

Answer 2

The electric field due to an infinitely large thin sheet with uniform surface charge density $\sigma$ is $E = \frac{|\sigma|}{2\epsilon_0}$, directed away from the sheet for positive $\sigma$ and towards the sheet for negative $\sigma$.

Configuration II, Region 3 Analysis:

In region 3 of configuration II, the electric field is calculated by summing the contributions from all five sheets. Sheets 1, 2, and 3 are to the left of region 3, while sheets 4 and 5 are to the right.

$E_{II,3} = \frac{\sigma_1'}{2\epsilon_0}(1) + \frac{\sigma_2'}{2\epsilon_0}(1) + \frac{\sigma_3'}{2\epsilon_0}(1) + \frac{\sigma_4'}{2\epsilon_0}(-1) + \frac{\sigma_5'}{2\epsilon_0}(-1)$

$E_{II,3} = \frac{1}{2\epsilon_0} (\frac{\sigma_0}{2} - \sigma_0 + \sigma_0 + \sigma_0 + \frac{\sigma_0}{2}) = \frac{1}{2\epsilon_0} (2\sigma_0) = \frac{\sigma_0}{\epsilon_0}$.

Therefore, the magnitude of the electric field in region 3 of configuration II is $\frac{\sigma_0}{\epsilon_0}$.