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

Step 1: Understanding the Concept:
When the permittivity of a dielectric varies along the thickness of a capacitor, we can treat it as a combination of infinitesimal capacitors in series.
The capacitance of a differential element of thickness \(dx\) is \(dC = \frac{\epsilon(x) A}{dx}\).
The total equivalent capacitance \(C_{eq}\) for elements in series is given by: \[\frac{1}{C_{eq}} = \int \frac{1}{dC} = \int_{0}^{d} \frac{dx}{\epsilon(x) A}\]
Step 2: Key Formula or Approach:
The total inverse capacitance is the sum of the inverse capacitances of the two halves due to symmetry.
\[\frac{1}{C} = \frac{1}{A} \left[ \int_{0}^{d/2} \frac{dx}{\epsilon_0 + kx} + \int_{d/2}^{d} \frac{dx}{\epsilon_0 + k(d-x)} \right]\]
Step 3: Detailed Explanation:
Let's solve the first integral:
\[I_1 = \int_{0}^{d/2} \frac{dx}{\epsilon_0 + kx} = \frac{1}{k} \left[ \ln(\epsilon_0 + kx) \right]_{0}^{d/2} = \frac{1}{k} \ln\left( \frac{\epsilon_0 + kd/2}{\epsilon_0} \right) = \frac{1}{k} \ln\left( \frac{2\epsilon_0 + kd}{2\epsilon_0} \right)\]
Now, solve the second integral:
Let \(u = d - x\), then \(du = -dx\). As \(x \to d/2, u \to d/2\); as \(x \to d, u \to 0\).
\[I_2 = \int_{d/2}^{d} \frac{dx}{\epsilon_0 + k(d-x)} = \int_{d/2}^{0} \frac{-du}{\epsilon_0 + ku} = \int_{0}^{d/2} \frac{du}{\epsilon_0 + ku} = I_1\]
The total inverse capacitance is:
\[\frac{1}{C} = \frac{1}{A} (I_1 + I_2) = \frac{2 I_1}{A} = \frac{2}{kA} \ln\left( \frac{2\epsilon_0 + kd}{2\epsilon_0} \right)\]
Taking the reciprocal to find \(C\):
\[C = \frac{kA}{2 \ln \left( \frac{2\epsilon_0 + kd}{2\epsilon_0} \right)}\]
Step 4: Final Answer:
The capacitance of the capacitor is \(\frac{kA}{2 \ln \left( \frac{2\epsilon_0 + kd}{2\epsilon_0} \right)}\).