Step 1: Recall concept

If a parallel-plate capacitor has partial dielectric filling in the direction of separation (i.e. layered dielectrics along thickness), then it is equivalent to capacitors in series.

Plate area: $A = 50 \,\text{cm}^2 = 50 \times 10^{-4}\, \text{m}^2 = 5 \times 10^{-3}\, \text{m}^2$.
Separation: $d = 3 \,\text{mm} = 3 \times 10^{-3}\,\text{m}$.
Thickness of dielectric: $t_d = 1 \times 10^{-3}\,\text{m}$. Relative permittivity: $\varepsilon_r = 4$.
Remaining thickness (air): $t_a = d - t_d = (3 - 1) \,\text{mm} = 2 \times 10^{-3}\,\text{m}$.

So effectively we have:

A capacitor with dielectric slab: $\; t_d = 1 \times 10^{-3}\,\text{m}, \;\varepsilon = \varepsilon_0 \cdot 4$.
Another capacitor with air gap: $\; t_a = 2 \times 10^{-3}\,\text{m}, \;\varepsilon = \varepsilon_0$.

Both share same area $A$ and are in series.

Step 2: Capacitance of each section

Dielectric part:

$ C_1 = \frac{\varepsilon_0 \varepsilon_r A}{t_d} = \frac{ 4 \varepsilon_0 \cdot 5\times 10^{-3}}{1 \times 10^{-3}} = \frac{20 \varepsilon_0}{1}\,. $

So, $C_1 = 20 \varepsilon_0$.

Air part:

$ C_2 = \frac{\varepsilon_0 A}{t_a} = \frac{ \varepsilon_0 \cdot 5\times 10^{-3}}{2\times 10^{-3}} = \frac{5}{2}\varepsilon_0 = 2.5 \varepsilon_0\,. $

Step 3: Equivalent capacitance (series combination)

$ \frac{1}{C_\text{eq}} = \frac{1}{C_1} + \frac{1}{C_2}. $

Substitute numbers:

$ \frac{1}{C_\text{eq}} = \frac{1}{20 \varepsilon_0} + \frac{1}{2.5 \varepsilon_0}. $

$ = \frac{1}{20 \varepsilon_0} + \frac{1}{(5/2)\varepsilon_0} = \frac{1}{20 \varepsilon_0} + \frac{2}{5 \varepsilon_0}. $

$ = \frac{1}{20 \varepsilon_0} + \frac{8}{20 \varepsilon_0} = \frac{9}{20 \varepsilon_0}. $

Hence,

$ C_\text{eq} = \frac{20}{9} \varepsilon_0. $

Final Answer:

$ \boxed{\tfrac{20 \varepsilon_0}{9}} $

That corresponds to Option B.

Capacitor

Continue this paper

Similar Capacitor PYQs

More practice from this question