Question

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

Hint:

For a parallel plate capacitor:  \[ C' = K \frac{C}{n} \]

where \(K\) is the dielectric constant, and \(n\) is the factor by which the plate separation increases.

Answer 1

Correct Answer: 105

The capacitance of a parallel plate capacitor is given by:

\[ C = \frac{\epsilon_0 A}{d} \]

where:

• \(\epsilon_0\) is the permittivity of free space,
• \(A\) is the area of the plates,
• \(d\) is the separation between the plates.

When a dielectric of constant \(K\) is introduced, the new capacitance becomes:

\[ C' = K \frac{\epsilon_0 A}{d} \]

In the given problem:

• Initial capacitance \(C = 15 \, \text{pF},\)
• The separation between the plates becomes twice (\(d \to 2d\)),
• A dielectric constant \(K = 3.5\) is introduced.

The new capacitance is:

\[ C' = K \frac{\epsilon_0 A}{2d} = \frac{K}{2} \times \frac{\epsilon_0 A}{d} = \frac{K}{2} \times C \]

Substitute \(C = 15 \, \text{pF}\) and \(K = 3.5\):

\[ C' = \frac{3.5}{2} \times 15 = \frac{52.5}{2} = 26.25 \, \text{pF} \]

The problem states that \(C' = \frac{x}{4} \, \text{pF}\). Equating:

\[ \frac{x}{4} = 26.25 \implies x = 26.25 \times 4 = 105 \]

Thus, the value of \(x\) is 105.

Answer 2

The correct answer is 105.
C0​=d∈0​A​=15pF 
C=2dK∈0​A​=23.5​×15pF=4105​pF