Question 7

MCQHARD

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 $\epsilon_0 = 9 \times 10^{-12} F/m$ )

(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 $\frac{\sigma_0}{\epsilon_0}$ . (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.

(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 $\frac{\sigma_0}{\epsilon_0}$ .

(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.

Detailed Solution

The electric field due to an infinite non-conducting sheet is $E = \frac{\sigma}{2\epsilon_0}$ . The net field in any region is $E = \frac{1}{2\epsilon_0}(\sum \sigma_{left} - \sum \sigma_{right})$ .

Given $\frac{\sigma_0}{\epsilon_0} = \frac{9 \times 10^{-6}}{9 \times 10^{-12}} = 10^6 V/m$ and $d = 10^{-6} m$ . (A) Config I: $\sigma$ values are $+\sigma_0, -\sigma_0, +\sigma_0, -\sigma_0, +\sigma_0, -\sigma_0$ . In region 4 (between sheets 4 and 5), $\sum \sigma_{left} = \sigma_0 - \sigma_0 + \sigma_0 - \sigma_0 = 0$ and $\sum \sigma_{right} = \sigma_0 - \sigma_0 = 0$ . So $E_4 = 0$ . (A is correct).

(B) Config II: $\sigma$ values are $+\frac{\sigma_0}{2}, -\sigma_0, +\sigma_0, -\sigma_0, +\sigma_0, -\frac{\sigma_0}{2}$ . In region 3, $\sum \sigma_{left} = \frac{\sigma_0}{2}$ and $\sum \sigma_{right} = -\frac{\sigma_0}{2}$ . $E_3 = \frac{1}{2\epsilon_0}(\frac{\sigma_0}{2} - (-\frac{\sigma_0}{2})) = \frac{\sigma_0}{2\epsilon_0}$ . Statement B says $\frac{\sigma_0}{\epsilon_0}$ , so B is incorrect.

(C) In Config I, fields in regions 1, 3, 5 are $\frac{\sigma_0}{\epsilon_0}$ and in regions 2, 4 are $0$ . $\Delta V = 3 \times \frac{\sigma_0}{\epsilon_0} \times d = 3 \times 10^6 \times 10^{-6} = 3 V$ . (C is incorrect).

(D) In Config II, fields in regions 1, 3, 5 are $\frac{\sigma_0}{2\epsilon_0}$ and in regions 2, 4 are $-\frac{\sigma_0}{2\epsilon_0}$ .

Total $\Delta V = (\frac{\sigma_0}{2\epsilon_0} - \frac{\sigma_0}{2\epsilon_0} + \frac{\sigma_0}{2\epsilon_0} - \frac{\sigma_0}{2\epsilon_0} + \frac{\sigma_0}{2\epsilon_0})d = \frac{\sigma_0}{2\epsilon_0}d = 0.5 V$ . (D is incorrect).

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