Question

A parallel plate condenser, with plate area A and distance between plates d, is filled with a medium whose permittivity varies as;

$\epsilon(x) = \epsilon_0 + \beta\epsilon_0x/d \quad 0 < x < \frac{d}{2}$

$\epsilon(x) = \epsilon_0 + \beta\epsilon_0\frac{(d-x)}{d} \quad \frac{d}{2} < x < d$

$x$ is the distance from one of the plates. For what value of $\beta$ would the capacity of the condenser be $\frac{1}{2\ln2}$ times that when it is completely filled with a uniform dielectric having dielectric constant $\beta$?

Answer 1

2

Answer 2

To solve this problem, we need to calculate the effective capacitance of the parallel plate capacitor with the given permittivity variation and then equate it to $\frac{1}{2\ln2}$ times the capacitance of a capacitor filled with a uniform dielectric of constant $\beta$. Here's a breakdown:

1. Effective Capacitance ($C_{eff}$):

   The capacitor can be thought of as consisting of infinitesimal capacitors in series. For an infinitesimal slab at position $x$, the capacitance $dC$ is given by:

   $dC = \frac{\epsilon(x) A}{dx}$

   The potential difference $dV$ contributed by this thickness $dx$ is:

   $dV = \frac{dq}{\epsilon(x) A}$

   For a series combination, the reciprocal of the effective capacitance is the integral of the reciprocal capacitances:

   $\frac{1}{C_{eff}} = \int_0^d \frac{dx}{\epsilon(x)A}$

2. Splitting the Integral:

   Since the permittivity function $\epsilon(x)$ is symmetric around $x = d/2$, we can split the integral into two identical parts:

   $\epsilon(x)=\epsilon_0\left(1+\frac{\beta x}{d}\right)$ for $0 < x < d/2$.

   The integral for one half is:

   $I = \int_0^{d/2} \frac{dx}{\epsilon_0\left(1+\frac{\beta x}{d}\right)} = \frac{d}{\epsilon_0\beta}\int_1^{1+\beta/2} \frac{du}{u} = \frac{d}{\epsilon_0\beta}\ln\left(1+\frac{\beta}{2}\right)$

   where $u = 1+\frac{\beta x}{d}$.

   Thus, the total integral is:

   $\frac{1}{C_{eff}} = 2I = \frac{2d}{\epsilon_0\beta}\ln\left(1+\frac{\beta}{2}\right)$

   Hence, the effective capacitance is:

   $C_{eff} = \frac{A\,\epsilon_0\beta}{2d\,\ln\left(1+\frac{\beta}{2}\right)}$

3. Uniform Dielectric Capacitance:

   For a capacitor completely filled with a uniform dielectric of dielectric constant $\beta$, the capacitance is:

   $C_{uniform} = \frac{\epsilon_0\beta A}{d}$

4. Applying the Given Condition:

   We are given that:

   $C_{eff} = \frac{1}{2\ln2}\,C_{uniform}$

   Substituting the expressions for $C_{eff}$ and $C_{uniform}$:

   $\frac{A\,\epsilon_0\beta}{2d\,\ln\left(1+\frac{\beta}{2}\right)} = \frac{1}{2\ln2}\left(\frac{\epsilon_0\beta A}{d}\right)$

   Canceling common factors $\frac{A\epsilon_0\beta}{d}$ (assuming $\beta \neq 0$):

   $\frac{1}{2\ln\left(1+\frac{\beta}{2}\right)} = \frac{1}{2\ln2}$

5. Solving for $\beta$:

   This implies:

   $\ln\left(1+\frac{\beta}{2}\right)=\ln2 \implies 1+\frac{\beta}{2}=2$

   Thus,

   $\frac{\beta}{2}=1 \implies \beta=2$

Therefore, the value of $\beta$ for which the given condition holds is 2.