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 $\frac{\mathrm{x}}{4}\mathrm{pF}$ . The value of $\mathrm{x}$ is __________

A parallel plate capacitor with air as the medium has an initial capacitance of 15 pF. The plate separation is doubled, and the space between the plates is filled with a dielectric material having a constant K = 3.5. The capacitance is given as x/4 pF. Let's find the value of x.

Complete Solution:

The capacitance of a parallel plate capacitor is given by:

C = (ε₀ K A) / d

• C: Capacitance
• ε₀: Permittivity of free space
• K: Dielectric constant
• A: Area of the plates
• d: Distance between the plates

Given Data

• Initial capacitance, C₁ = 15 pF
• Dielectric constant, K = 3.5
• New plate separation, d₂ = 2 × d₁
• New capacitance, C₂ = x / 4 pF

Step-by-Step Solution

When the dielectric constant changes and the distance doubles, the new capacitance is:

C₂ = (K / 2) × C₁

C₁ = 15 pF

K = 3.5

C₂ = (3.5 / 2) × 15 = 1.75 × 15 = 26.25 pF

The new capacitance is also given as:

C₂ = x / 4

Substituting C₂ = 26.25 pF:

26.25 = x / 4x = 26.25 × 4 = 105

Final Answer

The value of x is 105.