Question

The figure shows three infinite thin non-conducting charged plates perpendicular to the plane of the paper with charge per unit area +$\sigma$, +2$\sigma$ and –3$\sigma$. As we move from plate 1 to plate 2 magnitude of potential change occur is $V_{12}$, that from plate 2 to 3 is $V_{23}$ and that from plate 1 to 3 is $V_{13}$. Then-

• A. Ratio of net electric field at ponit A to that at point B is 1/3
• B. $9V_{12} = 2 V_{23}$
• C. $2V_{13} = 7 V_{12}$
• D. $3V_{13} = 5V_{23}$

Answer 1

Answer: (A)

Ratio of net electric field at ponit A to that at point B is 1/3

Answer 2

The problem involves calculating electric fields and potential differences due to three infinite thin non-conducting charged plates.

1. Electric Field due to a single infinite non-conducting charged plate:

The electric field magnitude due to a single infinite non-conducting charged plate with surface charge density $\sigma_s$ is given by $E = \frac{|\sigma_s|}{2\epsilon_0}$. The direction of the field is away from a positively charged plate and towards a negatively charged plate.

Let's define $E_0 = \frac{\sigma}{2\epsilon_0}$.

The plates are:

• Plate 1: $+\sigma$
• Plate 2: $+2\sigma$
• Plate 3: $-3\sigma$

Let's set up a coordinate system where Plate 1 is at $x=0$, Plate 2 at $x=2m$, and Plate 3 at $x=4m$. We'll consider the right direction as positive.

Electric field contributions from each plate:

• $E_1$: Due to Plate 1 ($+\sigma$). Field is $E_0$ (i.e., $\frac{\sigma}{2\epsilon_0}$) away from Plate 1.
• $E_2$: Due to Plate 2 ($+2\sigma$). Field is $2E_0$ (i.e., $\frac{2\sigma}{2\epsilon_0} = \frac{\sigma}{\epsilon_0}$) away from Plate 2.
• $E_3$: Due to Plate 3 ($-3\sigma$). Field is $3E_0$ (i.e., $\frac{3\sigma}{2\epsilon_0}$) towards Plate 3.

Net Electric Field in different regions:

• Region I (x < 0, Left of Plate 1):

  • $E_1$: Left ($-E_0$)
  • $E_2$: Left ($-2E_0$)
  • $E_3$: Right ($+3E_0$)
  • $E_{net, I} = -E_0 - 2E_0 + 3E_0 = 0$

• Region II (0 < x < 2m, Between Plate 1 and Plate 2, where Point A is):

  • $E_1$: Right ($+E_0$)
  • $E_2$: Left ($-2E_0$)
  • $E_3$: Right ($+3E_0$)
  • $E_{net, II} = +E_0 - 2E_0 + 3E_0 = 2E_0 = \frac{\sigma}{\epsilon_0}$ (rightward)
  • Magnitude of electric field at A: $|E_A| = \frac{\sigma}{\epsilon_0}$.

• Region III (2m < x < 4m, Between Plate 2 and Plate 3, where Point B is):

  • $E_1$: Right ($+E_0$)
  • $E_2$: Right ($+2E_0$)
  • $E_3$: Right ($+3E_0$)
  • $E_{net, III} = +E_0 + 2E_0 + 3E_0 = 6E_0 = \frac{3\sigma}{\epsilon_0}$ (rightward)
  • Magnitude of electric field at B: $|E_B| = \frac{3\sigma}{\epsilon_0}$.

• Region IV (x > 4m, Right of Plate 3):

  • $E_1$: Right ($+E_0$)
  • $E_2$: Right ($+2E_0$)
  • $E_3$: Left ($-3E_0$)
  • $E_{net, IV} = +E_0 + 2E_0 - 3E_0 = 0$

2. Evaluate Option (A): Ratio of net electric field at point A to that at point B.

$|E_A| = \frac{\sigma}{\epsilon_0}$

$|E_B| = \frac{3\sigma}{\epsilon_0}$

Ratio $\frac{|E_A|}{|E_B|} = \frac{\sigma/\epsilon_0}{3\sigma/\epsilon_0} = \frac{1}{3}$.

So, Option (A) is correct.

3. Calculate Potential Changes:

The potential difference between two points is given by $\Delta V = V_{final} - V_{initial} = -\int_{initial}^{final} \vec{E} \cdot d\vec{l}$. For a uniform electric field in a region, $\Delta V = -E_x \Delta x$.

• Magnitude of potential change from Plate 1 to Plate 2 ($V_{12}$):

  • Movement is from $x=0$ to $x=2m$. The electric field in this region (Region II) is $E_{net, II} = \frac{\sigma}{\epsilon_0}$ (rightward).
  • $V_2 - V_1 = -E_{net, II} \times (2m) = -\frac{\sigma}{\epsilon_0} \times 2 = -\frac{2\sigma}{\epsilon_0}$.
  • $V_{12} = |V_2 - V_1| = \frac{2\sigma}{\epsilon_0}$.

• Magnitude of potential change from Plate 2 to Plate 3 ($V_{23}$):

  • Movement is from $x=2m$ to $x=4m$. The electric field in this region (Region III) is $E_{net, III} = \frac{3\sigma}{\epsilon_0}$ (rightward).
  • $V_3 - V_2 = -E_{net, III} \times (2m) = -\frac{3\sigma}{\epsilon_0} \times 2 = -\frac{6\sigma}{\epsilon_0}$.
  • $V_{23} = |V_3 - V_2| = \frac{6\sigma}{\epsilon_0}$.

• Magnitude of potential change from Plate 1 to Plate 3 ($V_{13}$):

  • This is the sum of potential changes from 1 to 2 and 2 to 3.
  • $V_3 - V_1 = (V_3 - V_2) + (V_2 - V_1) = -\frac{6\sigma}{\epsilon_0} - \frac{2\sigma}{\epsilon_0} = -\frac{8\sigma}{\epsilon_0}$.
  • $V_{13} = |V_3 - V_1| = \frac{8\sigma}{\epsilon_0}$.

4. Evaluate Options (B), (C), (D):

• Option (B): $9V_{12} = 2 V_{23}$

  • LHS: $9V_{12} = 9 \times \frac{2\sigma}{\epsilon_0} = \frac{18\sigma}{\epsilon_0}$
  • RHS: $2V_{23} = 2 \times \frac{6\sigma}{\epsilon_0} = \frac{12\sigma}{\epsilon_0}$
  • $18 \neq 12$. So, Option (B) is incorrect.

• Option (C): $2V_{13} = 7 V_{12}$

  • LHS: $2V_{13} = 2 \times \frac{8\sigma}{\epsilon_0} = \frac{16\sigma}{\epsilon_0}$
  • RHS: $7V_{12} = 7 \times \frac{2\sigma}{\epsilon_0} = \frac{14\sigma}{\epsilon_0}$
  • $16 \neq 14$. So, Option (C) is incorrect.

• Option (D): $3V_{13} = 5V_{23}$

  • LHS: $3V_{13} = 3 \times \frac{8\sigma}{\epsilon_0} = \frac{24\sigma}{\epsilon_0}$
  • RHS: $5V_{23} = 5 \times \frac{6\sigma}{\epsilon_0} = \frac{30\sigma}{\epsilon_0}$
  • $24 \neq 30$. So, Option (D) is incorrect.

Based on the calculations, only Option (A) is correct.