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  4. A slab of material of dielectric constant K has the same area as the plates of a parallel plate capacitor but has a thickness ((3/4)) d, where d is the separation of the plates. The ratio of the capacitance C (in the presence of the dielectric) to the capacitance C0 (in the absence of the dielectric) is

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Q. A slab of material of dielectric constant $K$ has the same area as the plates of a parallel plate capacitor but has a thickness $\left(\frac{3}{4}\right) d$, where $d$ is the separation of the plates. The ratio of the capacitance $C$ (in the presence of the dielectric) to the capacitance $C_{0}$ (in the absence of the dielectric) is

AMUAMU 2010Electrostatic Potential and Capacitance

Solution:

The capacitance of a parallel plate capacitor in the absence of the dielectric is
$C_{0}=\frac{\varepsilon_{0} A}{d}$
If a dielectric slab is partially filled between the plates of capacitor
$C=\frac{\varepsilon_{0} A}{d-t+\frac{t}{K}}$
$C=\frac{\varepsilon_{0} A}{\left(d-\frac{3}{4} d\right)+\left(\frac{3 d}{4 K}\right)}$
$C =\frac{\varepsilon_{0} A}{\frac{d}{4}+\frac{3 d}{4 K}}$
$=\frac{4 K \varepsilon_{0} A}{d(K+3)}$
$\therefore \frac{C}{C_{0}} =\frac{4 K \varepsilon_{0} A}{d(K+3)} \times \frac{d}{\varepsilon_{0} A}$
$=\frac{4 K}{(K+3)}$