GATE 2026 IN Q.50: solution and answer

Official answer: CSource: Official answer key, GATE 2026 IN (IIT Guwahati)

1. For a parallel plate capacitor with air between the plates, $C=\frac{{ϵ}_{0}A}{d}$. With the plate area fixed, $C\propto \frac{1}{d}$.

2. The initial gap is $d_1=0.5\text{mm}$. The gap is decreased by $0.1\text{mm}$, so $d_2=0.5-0.1=0.4\text{mm}$.

3. Initial capacitance: $C_1=\frac{8.854\times 10^{-12}\times 2\times 10^{-4}}{0.5\times 10^{-3}}\approx 3.54\text{pF}$, using $A=2{\text{cm}}^2=2\times 10^{-4}{\text{m}}^2$.

4. Final capacitance: $C_2=\frac{8.854\times 10^{-12}\times 2\times 10^{-4}}{0.4\times 10^{-3}}\approx 4.43\text{pF}$.

5. The ratio does not depend on ${ϵ}_0$ or $A$: $\frac{C_2}{C_1}=\frac{d_1}{d_2}=\frac{0.5}{0.4}=1.25$.

6. Percentage change: $\frac{C_2-C_1}{C_1}\times 100=(1.25-1)\times 100=25\%$, an increase. This is option (C).