MHTCET 11 Apr 2026 Shift 1 Paper

Concept:A dielectric slab placed partially between the plates reduces the effective separation, which increases the capacitance of the capacitor.Explanation:The total separation between the plates is $2 \mathrm{~mm}$ and the dielectric slab has a thickness of $1 \mathrm{~mm}$.So, the remaining air gap is:$$2 \mathrm{~mm} - 1 \mathrm{~mm} = 1 \mathrm{~mm}$$For a parallel plate capacitor partly filled with a dielectric, the effective separation is given by:$$d_{\text{eff}} = d_{\text{air}} + \frac{d_{\text{dielectric}}}{K}$$Convert the given area into $\mathrm{m}^2$:$$A = 40 \mathrm{~cm}^2 = 40 \times 10^{-4} \mathrm{~m}^2 = 4 \times 10^{-3} \mathrm{~m}^2$$Here, converting the thicknesses to metres:$$d_{\text{air}} = 1 \mathrm{~mm} = 10^{-3} \mathrm{~m}, \quad d_{\text{dielectric}} = 1 \mathrm{~mm} = 10^{-3} \mathrm{~m}$$And the dielectric constant is $K = 5$.Now compute the effective separation:$$d_{\text{eff}} = 10^{-3} + \frac{10^{-3}}{5} = 10^{-3} + 0.2 \times 10^{-3} = 1.2 \times 10^{-3} \mathrm{~m}$$The capacitance of the system is:$$C = \frac{\epsilon_0 A}{d_{\text{eff}}} = \frac{\epsilon_0 \times 4 \times 10^{-3}}{1.2 \times 10^{-3}}$$$$C = \frac{4}{1.2} \epsilon_0 = \frac{10}{3} \epsilon_0 \mathrm{~F}$$Answer:The capacitance is $\frac{10}{3} \epsilon_0 \mathrm{~F}$, which matches Option C.