JEE Challenger
More from Electrostatic Potential and Capacitance
PhyMedAdv 2014 • Paper 1

Parallel Plate Capacitor with Partial Dielectric Slab Insertion

Consider a parallel plate capacitor containing a dielectric slab with dielectric constant KK covering 1/31/3 of its plate area, as illustrated in the figure. Let CC represent the overall capacitance of the capacitor and C1C_1 represent the capacitance of the region occupied by the dielectric. Upon charging the capacitor, the region with dielectric acquires a charge Q1Q_1, while the remaining area holds a charge Q2Q_2. The electric field inside the dielectric region is E1E_1, and in the dielectric-free region it is E2E_2. Ignoring edge effects, select the correct statement(s).

Question Diagram 1

Options

A

E1E2=1\frac{E_1}{E_2} = 1

Correct
B

E1E2=1K\frac{E_1}{E_2} = \frac{1}{K}

C

Q1Q2=3K\frac{Q_1}{Q_2} = \frac{3}{K}

D

CC1=2+KK\frac{C}{C_1} = \frac{2+K}{K}

Correct

Topics & Concepts

Electrostatic Potential and CapacitanceCapacitorsDielectricsElectric Field

Step-by-Step Solution

To determine the correct options, we analyze the setup of the parallel plate capacitor with a partially inserted dielectric slab.

1. Equivalent Capacitor Circuit Model

The dielectric slab of dielectric constant KK covers a fraction of 13\frac{1}{3} of the total plate area AA and extends through the entire separation distance dd between the parallel plates. This setup can be modeled as two individual capacitors connected in parallel across the same potential difference VV:

  • Region 1 (with dielectric):

    • Area: A1=A3A_1 = \frac{A}{3}
    • Distance between plates: dd
    • Dielectric constant: KK
    • Capacitance C1C_1: C1=Kε0A1d=Kε0A3dC_1 = \frac{K \varepsilon_0 A_1}{d} = \frac{K \varepsilon_0 A}{3d}

  • Region 2 (dielectric-free/vacuum):

    • Area: A2=2A3A_2 = \frac{2A}{3}
    • Distance between plates: dd
    • Dielectric constant: 11
    • Capacitance C2C_2: C2=ε0A2d=2ε0A3dC_2 = \frac{\varepsilon_0 A_2}{d} = \frac{2 \varepsilon_0 A}{3d}

2. Electric Field Analysis

Since both regions share the same potential difference VV across the common distance dd, the uniform electric field in each region is given by: E1=VdE_1 = \frac{V}{d} E2=VdE_2 = \frac{V}{d}

Taking the ratio of the electric fields: E1E2=V/dV/d=1\frac{E_1}{E_2} = \frac{V/d}{V/d} = 1

Thus, Option (A) is correct and Option (B) is incorrect.

3. Total Capacitance and Ratio CC1\frac{C}{C_1}

The total capacitance CC of the parallel combination is the sum of the individual capacitances: C=C1+C2=Kε0A3d+2ε0A3d=(K+2)ε0A3dC = C_1 + C_2 = \frac{K \varepsilon_0 A}{3d} + \frac{2 \varepsilon_0 A}{3d} = \frac{(K + 2) \varepsilon_0 A}{3d}

Taking the ratio of total capacitance CC to the dielectric portion capacitance C1C_1: CC1=(K+2)ε0A3dKε0A3d=2+KK\frac{C}{C_1} = \frac{\frac{(K + 2) \varepsilon_0 A}{3d}}{\frac{K \varepsilon_0 A}{3d}} = \frac{2 + K}{K}

Thus, Option (D) is correct.

4. Charge Distribution and Ratio Q1Q2\frac{Q_1}{Q_2}

The charges acquired by each portion under potential difference VV are: Q1=C1V=(Kε0A3d)VQ_1 = C_1 V = \left(\frac{K \varepsilon_0 A}{3d}\right) V Q2=C2V=(2ε0A3d)VQ_2 = C_2 V = \left(\frac{2 \varepsilon_0 A}{3d}\right) V

Taking the ratio of the charges: Q1Q2=C1C2=K2\frac{Q_1}{Q_2} = \frac{C_1}{C_2} = \frac{K}{2}

Thus, Option (C) is incorrect.

Conclusion

regions with the same potential difference yield E1E2=1\frac{E_1}{E_2} = 1 and CC1=2+KK\frac{C}{C_1} = \frac{2+K}{K}.

Correct Options: A and D

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