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Capacitance

NEET · Physics · Capacitance

A parallel plate capacitor with plate area A and separation d has...

A parallel plate capacitor with plate area AA and separation dd has an initial capacitance C0C_0 when there is air between the plates. It is then filled with three different dielectric materials. The left half of the space (thickness d/2d/2, full area AA) is filled with a dielectric of constant K1=2K_1 = 2. The right half of the space (thickness d/2d/2) is divided into two equal parts, each of area A/2A/2, and filled with dielectrics of constants K2=3K_2 = 3 and K3=5K_3 = 5. The new equivalent capacitance of the system is:

Options

  1. A83C0\dfrac{8}{3} C_0
  2. B85C0\dfrac{8}{5} C_0
  3. C478C0\dfrac{47}{8} C_0
  4. D103C0\dfrac{10}{3} C_0

Correct answer

A. 83C0\dfrac{8}{3} C_0

Step-by-step solution

The initial capacitance with air is C0=ε0AdC_0 = \dfrac{\varepsilon_0 A}{d}.

The system can be modeled as a capacitor C1C_1 in series with the parallel combination of capacitors C2C_2 and C3C_3.

For the left half (thickness d/2d/2, area AA):
C1=K1ε0Ad/2=2K1(ε0Ad)=2(2)C0=4C0C_1 = \dfrac{K_1 \varepsilon_0 A}{d/2} = 2K_1 \left(\dfrac{\varepsilon_0 A}{d}\right) = 2(2)C_0 = 4C_0.

For the right half, it is split into two equal areas (A/2A/2) with thickness d/2d/2:
C2=K2ε0(A/2)d/2=K2(ε0Ad)=3C0C_2 = \dfrac{K_2 \varepsilon_0 (A/2)}{d/2} = K_2 \left(\dfrac{\varepsilon_0 A}{d}\right) = 3C_0.
C3=K3ε0(A/2)d/2=K3(ε0Ad)=5C0C_3 = \dfrac{K_3 \varepsilon_0 (A/2)}{d/2} = K_3 \left(\dfrac{\varepsilon_0 A}{d}\right) = 5C_0.

C2C_2 and C3C_3 are in parallel, so their equivalent capacitance is:
Cp=C2+C3=3C0+5C0=8C0C_p = C_2 + C_3 = 3C_0 + 5C_0 = 8C_0.

Now, C1C_1 is in series with CpC_p. The total equivalent capacitance CeqC_{eq} is:
Ceq=C1CpC1+Cp=4C0×8C04C0+8C0=32C012=83C0C_{eq} = \dfrac{C_1 C_p}{C_1 + C_p} = \dfrac{4C_0 \times 8C_0}{4C_0 + 8C_0} = \dfrac{32C_0}{12} = \dfrac{8}{3}C_0.

Answer: 83C0\dfrac{8}{3} C_0
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