There are four capacitors: $C_1, C_2, C_3$ and $C_4$ with equal capacitance $C_0$-

(A) for capacitor $C_1$ area of the plates is doubled
(B) for capacitor $C_2$ distance between the plates is tripled
(C) the capacitor, $C_3$ is filled completely with a material having the dielectric constant $K = 3$
(D) for capacitor $C_4$ there is no change
Now, if the capacitors are arranged in decreasing order of their capacitances, then

Choose the correct answer from the options given below:

(C), (A), (D), (B)

The correct answer is Option (2) → (C), (A), (D), (B)

Capacitance of a parallel-plate capacitor: $C = \frac{\epsilon A}{d}$

Given initial capacitance $C_0$ for all capacitors.

Capacitor modifications:

(A) $C_1$: Area doubled, $C_1 = \frac{\epsilon (2A)}{d} = 2 C_0$

(B) $C_2$: Distance tripled, $C_2 = \frac{\epsilon A}{3d} = \frac{C_0}{3}$

(C) $C_3$: Filled with dielectric $K = 3$, $C_3 = \frac{3 \epsilon A}{d} = 3 C_0$

(D) $C_4$: No change, $C_4 = C_0$

Arrange in decreasing order of capacitance: $C_3 > C_1 > C_4 > C_2$

Final Answer: $C_3, C_1, C_4, C_2$