Physics Solution – Questions 22 & 23

Solution to Questions 22 & 23

Figure 1: Schematic of the two-layer capacitor connected to a DC source.

Solution to Question 22

1. Understanding the Steady State Condition

When a capacitor filled with conducting materials (leaky dielectrics) is connected to a DC voltage source, a current flows through it. In the steady state, charge does not accumulate indefinitely at the interface. This implies that the current flowing into the interface from Material A must equal the current flowing out into Material B.

Let $J$ be the current density (Current per unit area). The condition for steady current flow is:

$$ J_A = J_B $$

2. Applying Ohm’s Law and Kirchhoff’s Law

Using the point form of Ohm’s Law, $J = \sigma E$, where $\sigma$ is conductivity and $E$ is the electric field:

$$ \sigma_1 E_A = \sigma_2 E_B \implies E_B = \frac{\sigma_1}{\sigma_2} E_A \quad \dots(1) $$

The total potential difference $V$ across the plates is the sum of the potential drops across each layer:

$$ V = E_A d_1 + E_B d_2 $$

Substituting equation (1) for $E_B$:

$$ V = E_A d_1 + \left( \frac{\sigma_1}{\sigma_2} E_A \right) d_2 $$ $$ V = E_A \left( d_1 + \frac{\sigma_1 d_2}{\sigma_2} \right) $$ $$ V = E_A \left( \frac{\sigma_2 d_1 + \sigma_1 d_2}{\sigma_2} \right) $$

3. Solving for Electric Field in Material A

Rearranging the expression to solve for $E_A$:

$$ E_A = \frac{V \sigma_2}{d_1 \sigma_2 + d_2 \sigma_1} $$

Comparing this with the given options, the correct answer matches Option (d).

Solution to Question 23

1. Boundary Condition for Electric Fields

The question asks for the total surface charge density ($\sigma_{\text{total}}$) on the interface. This charge is responsible for the discontinuity in the electric field across the boundary.

Applying Gauss’s Law at the interface (pillbox method), the relationship between the normal components of the electric field and the total surface charge density is:

$$ E_{B} – E_{A} = \frac{\sigma_{\text{total}}}{\epsilon_0} $$

Note: We assume the field points from the positive plate (top) to the negative plate (bottom). The normal to the interface points downwards, from A to B.

2. Calculation

From Question 22, we have:

$$ E_A = \frac{V \sigma_2}{d_1 \sigma_2 + d_2 \sigma_1} $$

Using the relation $E_B = \frac{\sigma_1}{\sigma_2} E_A$:

$$ E_B = \frac{\sigma_1}{\sigma_2} \left( \frac{V \sigma_2}{d_1 \sigma_2 + d_2 \sigma_1} \right) = \frac{V \sigma_1}{d_1 \sigma_2 + d_2 \sigma_1} $$

Now, substituting these into the boundary condition equation:

$$ \sigma_{\text{total}} = \epsilon_0 (E_B – E_A) $$ $$ \sigma_{\text{total}} = \epsilon_0 \left( \frac{V \sigma_1}{d_1 \sigma_2 + d_2 \sigma_1} – \frac{V \sigma_2}{d_1 \sigma_2 + d_2 \sigma_1} \right) $$ $$ \sigma_{\text{total}} = \frac{\epsilon_0 V (\sigma_1 – \sigma_2)}{d_1 \sigma_2 + d_2 \sigma_1} $$

This result corresponds to Option (b).