Six infinitely large and thin non-conducting sheets are fixed in...

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 \mathrm{~m}

. The various regions between the sheets are denoted as

1,2,3,4

and 5. If

\sigma_0=9 \mu \mathrm{C} / \mathrm{m}^2

, then which of the following statements is/are correct: (Take permittivity of free space

\epsilon_0=9 \times 10^{-12} \mathrm{~F} / \mathrm{m}

Six infinitely large and thin non-conducting sheets are fixed in configurations I and II. As shown in the figure, the

Options

AIn region 4 of the configuration $I$ , the magnitude of the electric field is zero.
BIn region 3 of the configuration II, the magnitude of the electric field is $\frac{\sigma_0}{\epsilon_0}$ .
CPotential difference between the first and the last sheets of the configuration I is 5 V .
DPotential difference between the first and the last sheets of the configuration II is zero.

Correct answer

A. In region 4 of the configuration $I$ , the magnitude of the electric field is zero.

Step-by-step solution

\begin{aligned}\left(\mathrm{E}_4\right)_{\mathrm{I}}= & \frac{\sigma_0}{2 \epsilon_0}[1-1+1-1-1+1]=0 \\ \left(\mathrm{ V}_{\text{First }}\right)_{\mathrm{I}} & =\frac{-\sigma_0}{2 \epsilon_0}[-1+2-3+4-5] \mathrm{d} \\ & =\frac{-\sigma_0}{2 \epsilon_0}[-3] \mathrm{d}=\frac{\sigma_0 3 \mathrm{ d}}{2 \epsilon_0}\end{aligned}

\begin{aligned} & \begin{aligned}\left(V_{\text{Last }}\right)_I & =\frac{-\sigma_0}{2 \epsilon_0}[1-2+3-4+5] \\ & =\frac{\sigma_0}{2 \epsilon_0}[-3 \mathrm{ d}]\end{aligned}

\\ &

\begin{aligned}\left(V_{\text{First }}-V_{\text{Last }}\right)_I & =\frac{3 \sigma_0 \mathrm{ d}}{\epsilon_0} \\ & =\frac{3 \times 9 \times 10^{-6} \times 1 \times 10^{-6}}{9 \times 10^{-12}}=3 \mathrm{volt}\end{aligned}

\end{aligned}

\begin{aligned} & \begin{aligned}\left(\mathrm{E}_3\right)_{\text{II }}= & \frac{\sigma_0}{2 \epsilon_0}\left[\frac{1}{2}-1+1+1-1+\frac{1}{2}\right]=\frac{\sigma_0}{2 \epsilon_0} \\ & =\frac{-\sigma_0}{2 \epsilon_0}\left[-1+2-3+4-\frac{5}{2}\right] \mathrm{d} \\ & =\frac{-\sigma_0}{2 \epsilon_0}[2-2.5] \mathrm{d}=\frac{\sigma_0 \mathrm{ d}}{4 \epsilon_0}\end{aligned}

\\ &

\begin{aligned}\left(\mathrm{V}_{\text{Last }}\right)_{\text{II }} & =\frac{-\sigma_0}{2 \epsilon_0}\left[1-2+3-4+\frac{5}{2}\right] \mathrm{d} \\ & =\frac{-\sigma_0}{2 \epsilon_0}[6.5-6] \mathrm{d}=\frac{-\sigma_0 \mathrm{ d}}{4 \epsilon_0}\end{aligned}

\\ & \left(\mathrm{ V}_{\text{First }}-\mathrm{V}_{\text{Last }}\right)_{\text{II }}=\frac{\sigma_0 \mathrm{ d}}{2 \epsilon_0} \neq 0\end{aligned}$

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