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 σ0\sigma_0. The separation between any two consecutive sheets is 1μm1 \mu m. The various regions between the sheets are denoted as 1, 2, 3, 4 and 5. If σ0=9μC/m2\sigma_0 = 9 \mu C/m^2, then which of the following statements is/are correct: (Take permittivity of free space ϵ0=9×10−12F/m\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 σ0ϵ0\frac{\sigma_0}{\epsilon_0}. (C) Potential difference between the first and the last sheets of the configuration I is 5V5 V. (D) Potential difference between the first and the last sheets of the configuration II is zero.

Question
(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 σ0ϵ0\frac{\sigma_0}{\epsilon_0}.

(C)

Potential difference between the first and the last sheets of the configuration I is 5V5 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=σ2ϵ0E = \frac{\sigma}{2\epsilon_0}. The net field in any region is E=12ϵ0(∑σleft−∑σright)E = \frac{1}{2\epsilon_0}(\sum \sigma_{left} - \sum \sigma_{right}).

Given σ0ϵ0=9×10−69×10−12=106V/m\frac{\sigma_0}{\epsilon_0} = \frac{9 \times 10^{-6}}{9 \times 10^{-12}} = 10^6 V/m and d=10−6md = 10^{-6} m. (A) Config I: σ\sigma values are +σ0,−σ0,+σ0,−σ0,+σ0,−σ0+\sigma_0, -\sigma_0, +\sigma_0, -\sigma_0, +\sigma_0, -\sigma_0. In region 4 (between sheets 4 and 5), ∑σleft=σ0−σ0+σ0−σ0=0\sum \sigma_{left} = \sigma_0 - \sigma_0 + \sigma_0 - \sigma_0 = 0 and ∑σright=σ0−σ0=0\sum \sigma_{right} = \sigma_0 - \sigma_0 = 0. So E4=0E_4 = 0. (A is correct).

(B) Config II: σ\sigma values are +σ02,−σ0,+σ0,−σ0,+σ0,−σ02+\frac{\sigma_0}{2}, -\sigma_0, +\sigma_0, -\sigma_0, +\sigma_0, -\frac{\sigma_0}{2}. In region 3, ∑σleft=σ02\sum \sigma_{left} = \frac{\sigma_0}{2} and ∑σright=−σ02\sum \sigma_{right} = -\frac{\sigma_0}{2}. E3=12ϵ0(σ02−(−σ02))=σ02ϵ0E_3 = \frac{1}{2\epsilon_0}(\frac{\sigma_0}{2} - (-\frac{\sigma_0}{2})) = \frac{\sigma_0}{2\epsilon_0}. Statement B says σ0ϵ0\frac{\sigma_0}{\epsilon_0}, so B is incorrect.

(C) In Config I, fields in regions 1, 3, 5 are σ0ϵ0\frac{\sigma_0}{\epsilon_0} and in regions 2, 4 are 00. ΔV=3×σ0ϵ0×d=3×106×10−6=3V\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 σ02ϵ0\frac{\sigma_0}{2\epsilon_0} and in regions 2, 4 are −σ02ϵ0-\frac{\sigma_0}{2\epsilon_0}.

Total ΔV=(σ02ϵ0−σ02ϵ0+σ02ϵ0−σ02ϵ0+σ02ϵ0)d=σ02ϵ0d=0.5V\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).

Free Exam

Boost Your Exam Preparation!

Move beyond just reading solutions. Access our comprehensive Test Series, original Mock Tests, and interactive learning modules. Many premium tests are completely free!

  • Original Mocks & Regular Test Series
  • Real NTA-like Interface with Analytics
  • Many Free Tests Available