Question

A parallel plate capacitor with air between the plate has a capacitance of \(15\) pF. The separation between the plates becomes twice and the space between them is filled with a medium of dielectric constant \(3.5\). Then the capacitance becomes \(x/4\) pF.
The value of \(x\) is

• A. \(105\)
• B. \(109\)
• C. \(111\)
• D. \(115\)

Hint:

\(C=\frac{K\varepsilon_0A}{d}\), so doubling \(d\) halves it and \(K\) multiplies it by 3.5.

Answer 1

The Correct Option is A

Step 1: Plan:
Apply the two changes as separate factors.

Step 2: Steps:
Dielectric multiplies $C$ by $3.5$: $15\times3.5 = 52.5$ pF. Doubling the gap halves it: $\frac{52.5}{2} = 26.25$ pF $= \frac{105}{4}$ pF. So $x = 105$.

Final Answer:
The value of $x$ is $105$, option (A). \[ \boxed{105} \]