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Physics
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Capacitance

NEET · Physics · Capacitance

The space between the plates of a parallel plate capacitor is...

The space between the plates of a parallel plate capacitor is completely filled with two dielectric slabs of equal thickness d2\dfrac{d}{2}. The first slab has a dielectric constant K1=3K_1 = 3 and the second slab has a dielectric constant K2=6K_2 = 6. If the original capacitance with air between the plates was C0C_0, what is the ratio of the new equivalent capacitance to the original capacitance?

Options

  1. A4:14:1
  2. B9:29:2
  3. C2:12:1
  4. D1:41:4

Correct answer

A. 4:14:1

Step-by-step solution

The original capacitance with air is C0=ϵ0AdC_0 = \dfrac{\epsilon_0 A}{d}.

The two dielectric slabs can be considered as two capacitors connected in series, each with thickness d2\dfrac{d}{2}.

Capacitance of the first part: C1=K1ϵ0Ad/2=2×3ϵ0Ad=6C0C_1 = \dfrac{K_1 \epsilon_0 A}{d/2} = \dfrac{2 \times 3 \epsilon_0 A}{d} = 6 C_0

Capacitance of the second part: C2=K2ϵ0Ad/2=2×6ϵ0Ad=12C0C_2 = \dfrac{K_2 \epsilon_0 A}{d/2} = \dfrac{2 \times 6 \epsilon_0 A}{d} = 12 C_0

The equivalent capacitance CeqC_{eq} for the series combination is given by:
1Ceq=1C1+1C2\dfrac{1}{C_{eq}} = \dfrac{1}{C_1} + \dfrac{1}{C_2}

Ceq=C1C2C1+C2=6C0×12C06C0+12C0=72C018=4C0C_{eq} = \dfrac{C_1 C_2}{C_1 + C_2} = \dfrac{6 C_0 \times 12 C_0}{6 C_0 + 12 C_0} = \dfrac{72 C_0}{18} = 4 C_0

The ratio of the new capacitance to the original capacitance is 4:14:1.

Answer: 4:14:1
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