Two identical capacitors are connected in parallel combination. Their equivalent capacitance is \( C_1 \). If the plate area is halved and separation between plates is doubled for both the capacitors, their equivalent capacitance now is \( C_2 \). Find the ratio \( \displaystyle\frac{C_1}{C_2} \)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 4:1 \)
The capacitance of a parallel plate capacitor is \( C = \varepsilon_0 A/d \), and capacitors in parallel simply add.
\( \displaystyle C_1 = C + C = 2C \)
With the plate area halved and the separation doubled, each capacitor becomes
\( \displaystyle C' = \frac{\varepsilon_0 (A/2)}{2d} = \frac{C}{4} \)
\( \displaystyle C_2 = 2C' = \frac{C}{2} \)
\( \displaystyle \frac{C_1}{C_2} = \frac{2C}{C/2} = 4 \)
So \( C_1 : C_2 = 4 : 1 \).