Question

A parallel plate capacitor has plates of area A separated by distance 'd' between them. It is filled with a dielectric which has a dielectric constant that varies as k(x) = K(1 + \( \alpha \)x) where 'x' is the distance measured from one of the plates. If (ad) << 1, the total capacitance of the system is best given by the expression :

• A. \( {{A{ \in _0}K} \over d}\left( {1 + {{\left( {{{\alpha d} \over 2}} \right)}^2}} \right) \)
• B. \( {{A{ \in _0}K} \over d}\left( {1 + {{\alpha d} \over 2}} \right) \)
• C. \( {{A{ \in _0}K} \over d}\left( {1 + {{{\alpha ^2}{d^2}} \over 2}} \right) \)
• D. \( {{A{ \in _0}K} \over d}\left( {1 + \alpha d} \right) \)

Hint:

Treat the dielectric as thin slices in series: \(1/C=\int dx/(\epsilon_0 k(x)A)\), then expand for \(\alpha d\ll1\).

Answer 1

The Correct Option is B

Thin slice of thickness dx at distance x has \(dC^{-1}=\dfrac{dx}{\epsilon_0k(x)A}\). Slices are in series:
\(\dfrac1C=\int_0^d\dfrac{dx}{\epsilon_0AK(1+\alpha x)}=\dfrac{1}{\epsilon_0AK\alpha}\ln(1+\alpha d)\).
For \(\alpha d\ll1\): \(\ln(1+\alpha d)\approx\alpha d-\dfrac{\alpha^2d^2}{2}\), so \(\dfrac1C\approx\dfrac{d}{\epsilon_0AK}\left(1-\dfrac{\alpha d}{2}\right)\).
\(C\approx\dfrac{\epsilon_0AK}{d}\left(1-\dfrac{\alpha d}{2}\right)^{-1}\approx\dfrac{A\epsilon_0K}{d}\left(1+\dfrac{\alpha d}{2}\right)\).
Answer: (B)