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: Understanding the Concept:
The capacitance of a parallel plate capacitor is \(C = \frac{K\varepsilon_0A}{d}\). Initially (air, \(K=1\)): \(C_0 = \frac{\varepsilon_0A}{d} = 15\) pF.

Step 2: New capacitance:
New separation is \(2d\) and \(K = 3.5\). \[ C = \frac{3.5\,\varepsilon_0A}{2d} = \frac{3.5}{2}C_0 = 1.75\times15 = 26.25\ \text{pF} \]

Step 3: Find x:
\(26.25 = \frac{x}{4}\), so \(x = 105\).

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