Capacitance of a Parallel Plate Capacitor with a Dielectric Slab
Step 1: Capacitance without dielectric
For a parallel plate capacitor with vacuum (or air) between plates:
$C_0 = \frac{\varepsilon0 A}{d} \quad ...(i)$
where
- (A) = area of each plate
- (d) = separation between plates
- ($\varepsilon_0$) = permittivity of free space
Step 2: Introducing dielectric slab
Let a dielectric slab of thickness (t) and dielectric constant (K) be inserted between the plates.
- Electric field in free space region (thickness (d - t)): ($E_0$)
- Electric field in dielectric region (thickness (t)): ($E = \frac{E_0}{K}$)
Step 3: Potential difference
Potential difference between plates:
$V = E_0(d - t) + \frac{E_0 t}{K} = E_0 \left[(d - t) + \frac{t}{K}\right] \quad ...(ii)$
Since
$E_0 = \frac{q}{A \varepsilon_0}$
we get
$V = \frac{q}{A \varepsilon_0} \left[(d - t) + \frac{t}{K}\right] \quad ...(iii)$
Step 4: Capacitance expression
By definition,
$C = \frac{q}{V} = \frac{\varepsilon_0 A}{(d - t) + \frac{t}{K}} \quad ...(v)$
Step 5: Relation with ($C_0$)
Divide equation (v) by equation (i):
$\frac{C}{C_0} = \frac{1}{\left(1 - \frac{t}{d}\right) + \frac{t}{dK}}$
$\Rightarrow C = \frac{C_0}{1 - \frac{t}{d}\left(1 - \frac{1}{K}\right)} \quad ...(vi)$
Since ($K = \varepsilon_r$) (relative permittivity):
$C = \frac{C_0}{1 - \frac{t}{d}\left(1 - \frac{1}{\varepsilon_r}\right)}$
Step 6: Special case
If the dielectric slab completely fills the space ((t = d)):
$C = KC_0 \quad ...(vii)$
Thus, capacitance increases by a factor equal to the dielectric constant.
Step 7: Definition of dielectric constant
From equation (vii):
$K = \frac{C}{C_0} \quad ...(viii)$
So, dielectric constant is defined as the ratio of capacitance with dielectric to capacitance without dielectric.
✅ Final Result:
$C = \frac{\varepsilon_0 A}{(d - t) + \frac{t}{K}} = \frac{C_0}{1 - \tfrac{t}{d}\left(1 - \tfrac{1}{K}\right)}$