Question

A parallel plate capacitor with plate area 'A' and distance of separation 'd' is filled with a dielectric. What is the capacity of the capacitor when permittivity of the dielectric varies as : \[ \epsilon(x) = \epsilon_0 + kx, \text{ for } \left( 0<x \le \frac{d}{2} \right) \] \[ \epsilon(x) = \epsilon_0 + k(d-x), \text{ for } \left( \frac{d}{2} \le x \le d \right) \]

• A. $\frac{kA}{2 \ln \left( \frac{2\epsilon_0 + kd}{2\epsilon_0} \right)}$
• B. $\frac{kA}{2} \ln \left( \frac{2\epsilon_0}{2\epsilon_0 - kd} \right)$
• C. $(\epsilon_0 + \frac{kd}{2})^{2/kA}$
• D. $0$

Hint:

For dielectrics varying in the direction of the field (perpendicular to plates), always use the series combination formula: \(1/C = \int dx / (\epsilon A)\).
If the dielectric varies parallel to the plates, use the parallel combination formula: \(C = \int \epsilon dA / d\).

Answer 1

The Correct Option is A

Given:

A parallel plate capacitor with dielectric whose permittivity varies as:

For 0 < x ≤ d/2:
ε(x) = ε₀ + kx

For d/2 ≤ x ≤ d:
ε(x) = ε₀ + k(d − x)

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Step 1: Capacitance of non-uniform dielectric

For an infinitesimal slice of thickness dx:

dC = ε(x)A / dx

Since slices are in series:

1/C = ∫ (dx / ε(x)A)

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Step 2: Contribution from first region (0 to d/2)

1/C₁ = ∫₀^{d/2} dx / [A(ε₀ + kx)]

1/C₁ = (1/kA) ln[(ε₀ + kd/2)/ε₀]

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Step 3: Contribution from second region (d/2 to d)

1/C₂ = ∫_{d/2}^{d} dx / [A(ε₀ + k(d − x))]

Substitute y = d − x:

1/C₂ = (1/kA) ln[(ε₀ + kd/2)/ε₀]

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Step 4: Total capacitance

1/C = 1/C₁ + 1/C₂

1/C = (2/kA) ln[(ε₀ + kd/2)/ε₀]

Therefore,

C = kA / [2 ln((ε₀ + kd/2)/ε₀)]

Rewriting:

C = kA / [2 ln((2ε₀ + kd)/(2ε₀))]

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Final Answer:

C = kA / [2 ln((2ε₀ + kd)/(2ε₀))]