Capacitor With Dielectrics

Practice Questions:

Question 1: difficult

A parallel plate condenser is filled with two dielectrics as shown in figure. Area of each plate is A metre² and the separation is d metre. The dielectric constants are K1 and K2 respectively. Its capacitance in farad will be :

1. \[\frac{\varepsilon_{0A}}{d}\left( K_{1} +K_{2}\right)\]

2. \[\frac{\varepsilon_{0A}}{d}\frac{\left( K_{1} +K_{2}\right)}{2}\]

3. \[\frac{\varepsilon_{0A}}{d}2\left( K_{1} +K_{2}\right)\]

4. \[\frac{\varepsilon_{0A}}{d}\frac{\left( K_{1} -K_{2}\right)}{2}\]

View Answer

Both the parts can be taken as separate capacitors connected in parallel.

So, C=C1+ C2=\(\frac{\varepsilon_{0A}}{d}\frac{\left( K_{1} +K_{2}\right)}{2}\)

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Question 2: difficult

The capacity of a parallel plate condenser is \(C_0\). If a dielectric of relative permittivity \(\varepsilon_r\) and thickness equal to one fourth the plate separation is placed between the plates, then its capacity becomes \(C\). The value of \(\frac{C}{C_0}\) will be :

1. \(\frac{5\varepsilon_r}{4\varepsilon_r + 1}\)

2. \(\frac{4\varepsilon_r}{3\varepsilon_r + 1}\)

3. \(\frac{3\varepsilon_r}{2\varepsilon_r + 1}\)

4. \(\frac{2\varepsilon_r}{\varepsilon_r + 1}\)

View Answer

The initial capacitance is \(C_0 = \frac{\varepsilon_0 A}{d}\). With a dielectric slab of thickness \(t = d/4\), the new capacitance is \(C = frac{\varepsilon_0 A}{d - t + \frac{t}{\varepsilon_r}} = \frac{\varepsilon_0 A}{\frac{3d}{4} + \frac{d}{4\varepsilon_r}}\). Simplifying this gives \(C = C_0 \frac{4\varepsilon_r}{3\varepsilon_r + 1}\).

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