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:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (C), (A), (D), (B)
The capacitance of a parallel plate capacitor is
\( \displaystyle C = \frac{K\varepsilon_0 A}{d} \)
so it is proportional to the area and to the dielectric constant, and inversely proportional to the separation. Starting from \( C_0 \) each time:
(A) for capacitor \( C_1 \) area of the plates is doubled — \( C_1 = 2C_0 \)
(B) for capacitor \( C_2 \) distance between the plates is tripled — \( C_2 = C_0/3 \)
(C) the capacitor, \( C_3 \) is filled completely with a material having the dielectric constant \( K = 3 \) — \( C_3 = 3C_0 \)
(D) for capacitor \( C_4 \) there is no change — \( C_4 = C_0 \)
\( 3C_0 \gt 2C_0 \gt C_0 \gt C_0/3 \)
The decreasing order is (C), (A), (D), (B).