JEE Physics 2023 Question with Solution

Given: Plate area $A = 40 \times 10^{-4} \, \text{m}^2$, total separation $d = 2 \, \text{mm}$, dielectric slab thickness $t = 1 \, \text{mm}$, dielectric constant $k = 5$.

Find: Equivalent capacitance of the system.

The arrangement is equivalent to two capacitors in series: one with dielectric and one with air.

For the dielectric-filled part,

For the air-filled part,

Substituting $A = 40 \times 10^{-4} \, \text{m}^2$, $t = 1 \times 10^{-3} \, \text{m}$, and $d-t = 1 \times 10^{-3} \, \text{m}$,

Now combine them in series:

Therefore, the capacitance of the system is $\frac{10}{3} \varepsilon_0 \, \text{F}$. The correct option is C.

The first solution block lists option A, but the worked calculation clearly gives $\frac{10}{3} \varepsilon_0 \, \text{F}$, so the solution working has been used as the authority.

Given: A dielectric fills only part of the plate separation.

Find: How to model the capacitor.

Since the dielectric occupies thickness $t$ and the remaining space of thickness $d-t$ is air, the electric field passes successively through the dielectric region and the air region. Hence these two regions behave like capacitors connected in series, both having the same plate area $A$.

Using this model directly leads to

Substituting $t = 1 \times 10^{-3} \, \text{m}$, $d-t = 1 \times 10^{-3} \, \text{m}$, $k = 5$, and $A = 40 \times 10^{-4} \, \text{m}^2$ gives the same result:

Therefore, the correct option is C.