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 initial capacitance \(C_0\) of a parallel plate capacitor with air between the plates is given by:

\(C_0 = 15\, pF\)

The formula for capacitance of a parallel plate capacitor is:

\(C = \frac{\varepsilon_0 \cdot A}{d}\)

where \(\varepsilon_0\) is the permittivity of free space, \(A\) is the area of the plates, and \(d\) is the distance between the plates.

When the distance between the plates is doubled, the capacitance becomes:

\(C' = \frac{\varepsilon_0 \cdot A}{2d} = \frac{C_0}{2}\)

If a dielectric with constant \(k = 3.5\) is introduced, the new capacitance becomes:

\(C'' = k \cdot \frac{\varepsilon_0 \cdot A}{2d} = k \cdot \frac{C_0}{2}\)

Calculating with \(k = 3.5\) and \(C_0 = 15\, pF\):

\(C'' = 3.5 \cdot \frac{15}{2} = 3.5 \cdot 7.5 = 26.25\, pF\)

The problem states \(C'' = \frac{x}{4} pF\):

Equating,

\(\frac{x}{4} = 26.25\)

Solving for \(x\):

\(x = 26.25 \times 4 = 105\)

The value of \(x\) is 105, which falls within the expected range \(105,105\).