Question

A slab of material of dielectric constant K has the same area as the plates of a parallel plate capacitor but has a thickness t (t<d). Find the ratio of the capacitance with dielectric inside it to its capacitance without the dielectric.

Answer 1

Where Given

k = Dielectric constant

t = thickness of dielectric

Here; the net electric field experienced by the dielectric is

$\overrightarrow{E}={\overrightarrow{E}}_0-{\overrightarrow{E}}_p$ where $\overrightarrow{E_0}$. ___(1)

the applied E. Field and $\overrightarrow{E_p}$ is the electric field that is produced inside the dielectric due to polarization of charges.

Potential ditlerence between the plates :

$V=E_0(d-t)+Et.$ __(2) $\because E=\frac{dv}{dx}$

But $E=\frac{E_0}{k}$ __(3)

from (2) and (3)

$\Rightarrow V=E_0[d-t+\frac{t}{k}]$ __(4)

$E_0=\frac{\epsilon }{{\epsilon }_0}=\frac{Q}{A{\epsilon }_0}$ __(5)

$V=\frac{Q}{A{\epsilon }_0}[d-t+\frac{t}{k}]$ __(6)

Since $Q=CV\Rightarrow C=\frac{V}{Q}$ __(7)

From (6) and (7)

$C_k=\frac{{\epsilon }_{0}A}{(d-t+\frac{t}{k})}$ and $C=\frac{{\epsilon }_{0}A}{d}$

Hence,

the ratio of the two:-

$\frac{Ck}{C}=\frac{\frac{{\epsilon }_{0}A}{(d-t+\frac{t}{k})}}{\frac{{\epsilon }_{0}A}{d}}$

$=\frac{d}{(d-t+\frac{t}{k})}$