Question

Derive an expression for the capacitance of a parallel-plate capacitor when the medium between the plates is partially filled with a dielectric medium. What will be the effect on capacitance if it is completely filled by metallic strip?

Hint:

Introducing a dielectric slab always increases the capacitance ($C>C_0$). The effective distance between the plates decreases from $d$ to $(d - t + t/K)$, which makes the denominator smaller and the capacitance larger.

Answer 1

Step 1: Setting up the Parameters.
Consider a parallel-plate capacitor with plate area $A$ and separation $d$. Let a dielectric slab of thickness $t$ ($t<d$) and dielectric constant $K$ be introduced between the plates. The electric field in the vacuum region is $E_0 = \frac{\sigma}{\epsilon_0}$, and the field inside the dielectric is $E = \frac{E_0}{K}$.

Step 2: Calculating Potential Difference.
The total potential difference $V$ between the plates is the sum of the potential across the air gap and the potential across the dielectric: $$V = E_0(d - t) + Et$$ $$V = E_0(d - t) + \frac{E_0}{K}t = E_0 \left[ (d - t) + \frac{t}{K} \right]$$ Substituting $E_0 = \frac{Q}{A\epsilon_0}$: $$V = \frac{Q}{A\epsilon_0} \left[ (d - t) + \frac{t}{K} \right]$$
Step 3: Deriving Capacitance.
Capacitance $C$ is defined as $C = \frac{Q}{V}$. Substituting the expression for $V$: $$C = \frac{Q}{\frac{Q}{A\epsilon_0} \left[ (d - t) + \frac{t}{K} \right]} = \frac{A\epsilon_0}{(d - t) + \frac{t}{K}}$$ This is the required expression.

Step 4: Effect of a Metallic Strip.
If the space is completely filled with a metallic strip, then $t = d$. For a conductor, the dielectric constant $K \rightarrow \infty$. Substituting these into the formula: $$C = \frac{A\epsilon_0}{(d - d) + \frac{d}{\infty}} = \frac{A\epsilon_0}{0} = \infty$$ The capacitance becomes infinite because the two plates are effectively short-circuited.