Question

The capacity of a parallel plate air capacitor is \[C_0\] and the distance of separation of the plates is d. A dielectric slab of thickness \[\frac{4\mathrm{d}}{5}\] is introduced between the plates and the new capacity is C. The ratio \[C:C_{0}\] is ______.

Options

• 5K: (K+4)

• 4 K: (K+5)

• (K+4): 5K

• 5K: 4

Answer 1

The capacity of a parallel plate air capacitor is \[C_0\] and the distance of separation of the plates is d. A dielectric slab of thickness \[\frac{4\mathrm{d}}{5}\] is introduced between the plates and the new capacity is C. The ratio \[C:C_{0}\] is 5K: (K+4).

Explanation:

For an air capacitor, \[\mathrm{C_{0}=\frac{A\varepsilon_{0}K}{d}=\frac{A\varepsilon_{0}}{d}}\]      ...(K = 1 for air)

When dielectric slab of thickness \[\frac{4\mathrm{d}}{5}\] is inserted the combination acts as capacitors in series.

\[\frac{1}{\mathrm{C}}=\frac{1}{\frac{\mathrm{A}\varepsilon_0\mathrm{K}_1}{\frac{\mathrm{d}}{5}}}+\frac{1}{\frac{\mathrm{A}\varepsilon_0\mathrm{K}_2}{\frac{4\mathrm{d}}{5}}}\]

       \[=\frac{\mathrm{d}}{5\mathrm{A}\varepsilon_0}\left(\frac{4+\mathrm{K}}{\mathrm{K}}\right)\quad\ldots(\mathrm{K}_1\]= 1. for air)

\[\mathrm{C}=\frac{5\mathrm{A}\varepsilon_\mathrm{o}}{\mathrm{d}}\left(\frac{\mathrm{K}}{4+\mathrm{K}}\right)\]

\[\frac{\mathrm{C}}{\mathrm{C}_0}=\frac{\frac{5\mathrm{A}\varepsilon_0}{\mathrm{d}}\left(\frac{\mathrm{K}}{4+\mathrm{K}}\right)}{\frac{\mathrm{A}\varepsilon_0}{\mathrm{d}}}\]

\[\frac{\mathrm{C}}{\mathrm{C}_0}=\left(\frac{5\mathrm{K}}{4+\mathrm{K}}\right)\]