Question

Explain about (capacitance parallel plate - dielectric slab).

Answer 1

Suppose +Q is the charge on one plate and -Q Is charge on the second plate. Bring a rectangular slab made up of conducting material between the plates of the capacitor. The thickness of the slab must be less than the distance between the plates of the capacitor. When the electric field will be applied then polarization of molecules will be started. The polarization will take place in me direction same as that of electric held. Consider a vector mat must be polarized. name It as P The polarization vector must be in the direction of electric field $E_0$. Then this vector will start Its functioning and will produce an electric field $E_0$ in the opposite direction to that of $E_0$. The net electric held in the circuit is shown by the figure.

$E=E_0-E_p$
The electric field $E_0$ in the outside region of the dielectric will be null. Now the equation of the potential difference between the plates will be
$V=(d-t)+Et$
But $E_0=E_{r}orK$
Therefore $E=\frac{E_0}{k}$
So, $V=E_0(d-t)+\frac{E_{0}t}{k}$
$V=E_0[d-t+\frac{t}{k}]$
As we know
$E_0=\frac{Q}{E_0}$
$=\frac{Q}{A{ϵ}_0}$
$V=\frac{Q}{A{ϵ}_0}[d-t+\frac{t}{k}]$
Capacitance of the capacitor is shown in the equation below:
$C=\frac{Q}{V}=\frac{A{ϵ}_0}{(d-t+\frac{t}{k})}$
$=\frac{{ϵ}_{0}A}{d-t(1-\frac{1}{k})}$
i.e., $C=\frac{{ϵ}_{0}A}{d-t(1-\frac{1}{k})}...2.31$
So, $C>C_0$
Clearly, it is proved that if a dielectric slab is placed in the plates of a capacitor then its capacitance will increase by some amount.