MHTCET 11 Apr 2026 Shift 2 Paper

Concept:For a parallel plate capacitor, capacitance is $$C=\frac{\epsilon_{0} A}{d}$$ When capacitors are connected in parallel, the equivalent capacitance is the sum of individual capacitances.Explanation:Each capacitor has plate area $\frac{A}{3}$, so substitute this area in the formula.For the first capacitor with separation $d$:$$C_{1}=\frac{\epsilon_{0}\left(\frac{A}{3}\right)}{d}=\frac{\epsilon_{0} A}{3 d}$$ For the second capacitor with separation $2d$:$$C_{2}=\frac{\epsilon_{0}\left(\frac{A}{3}\right)}{2d}=\frac{\epsilon_{0} A}{6 d}$$ For the third capacitor with separation $3d$:$$C_{3}=\frac{\epsilon_{0}\left(\frac{A}{3}\right)}{3d}=\frac{\epsilon_{0} A}{9 d}$$ In parallel combination, capacitances add directly:$$C_{eq}=C_{1}+C_{2}+C_{3}$$ $$C_{eq}=\frac{\epsilon_{0} A}{d}\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{9}\right)$$ Now, $\frac{1}{3}+\frac{1}{6}+\frac{1}{9}=\frac{6+3+2}{18}=\frac{11}{18}$.Therefore,$$C_{eq}=\frac{11 \epsilon_{0} A}{18 d}$$Answer:The equivalent capacitance is $\frac{11 \epsilon_{0} A}{18 d}$, which matches Option C.