Question

Two circular plates each of radius '\(r\)' are kept parallel to each other distance '\(d\)' apart. The capacitance of the capacitor formed is '\(C_1\)'. If the radius of each of the plates is increased to \(\sqrt{3}\) times the earlier radius and their distance of separation decreased to half the initial value, the capacitance now becomes '\(C_2\)'. The ratio \(C_1:C_2\) is

• A. \(1:2\)
• B. \(1:4\)
• C. \(1:6\)
• D. \(6:1\)

Hint:

Capacitance is proportional to area and inversely proportional to separation.

Answer 1

The Correct Option is C

Step 1: Track each change:
Area becomes $3$ times: $C$ goes up by $3$. Distance becomes $\frac12$: $C$ goes up by $2$.

Step 2: Combine:
The total factor is $3 \times 2 = 6$, so $C_2 = 6C_1$ and $C_1 : C_2 = 1 : 6$.

Final Answer:
The ratio is $1:6$, option (C). \[ \boxed{1:6} \]