Explain in detail the effect of a dielectric placed in a parallel plate capacitor.

Explanation

(i) When the capacitor is disconnected from the battery: Consider a capacitor with two parallel plates each of cross - sectional area A and are separated by a distance d. The capacitor is charged by a battery of voltage `V_(0)` and the charge stored is `Q_(0)` . The capacitance of the capacitor without the dielectric is
`C_(0)= (Q_(0))/(V_(0))`
The battery is then disconnected from the capacitor and the dielectric is inserted between the plates . The introduction of dielectric between the plates will decrease the electric field . Experimentally it is found that the modifed electric field is given by
`E = (E_(0))/(epsilon_(0))`

Here `E_(0)` is the electric field inside the capacitors when there is no dielectric and `epsilon_(r)` is the relative permeability of the dielectric or simply known as the dielectric constant . Since `epsilon_(r) gt 1` the electric field `E lt E_(0)` . As a result the electrostatic potential difference between the plates ( V - Ed) is also reduced . But at the same time the charge `Q_(0)` will remain constant one the battery is disconnected. Hence the new potential difference is
`V =Ed = (E_(0))/(epsilon_(0)) d = (V_(0))/(epsilon_(r))`
We know that capacitance is inversely proportional to the potential difference . Therefore as V decreases C increases . Thus new capacitance in the presence of a dielectric is
`C = (Q_(0))/(V) = epsilon_(0) (Q_(0))/(V_(0)) = epsilon_(r) C_(0)`
Since `e_(r) gt 1` we have `C gt C_(0)` . Thus insertion of the dielectric `epsilon_(r)` increases the capacitance . Using equation .
`C= (epsilon_(0)A)/(d) `
`C= (epsilon_(r)epsilon_(0)A)/(d) = (epsilonA)/(d)`
where `epsilon= epsilon_(r)epsilon_(0)` is the permittivity of the dielectric medium . The energy stored in the capacitor before the insertion of a dielectric is given by
`U_(0)=(1)/(2) (Q_(0)^(2))/(C_(0))`
After the dielectric is inserted the charge `Q_(0)` remains constant but the capacitance is increased . As a result the stored energy is decreased .
`U = (1)/(2) (Q_(0)^(2))/(2C) =(1)/(2) (Q_(0)^(2))/(2epsilon_(r)C_(0))= (U_(0))/(epsilon_(r))`
Since `epsilon_(r) gt 1 ` we get `U lt U_(0)` . There is a decrease in energy because when the dielectric is inserted the capacitor spends some energy in pulling the dielectric inside .
(ii) When the battery remains connceted to the capacitor : Let us now condsider what happens when the battery of voltage `V_(0)` remains connceted to the capacitor when the dielectric is inserted into the capacitor . The potential difference `V_(0)` across the plates remains constant . But it is found experimentally ( first shown by Faraday ) that when dielectric is inserted the charge stored in the capacitor is increased by a fatcor `epsilon_(r)` .
`Q= epsilon_(r) Q_(0)`
Due to this increased charge the capacitance is also increased . The new capacitance is
`C= (Q)/(V_(0))= epsilon_(r) (Q_(0))/(V_(0))= epsilon_(r)C_(0)`

However the reason for the increase in capacitance in this case when the battery remains connected is diferent from the case when the battery is disconnected before introducing the dielectric .
Now `C_(0)= (epsilon_(0)A)/(d) ` and `C= (epsilon_(0)A)/(d)`
The energy stored in the capacitor before the insertion of a dielectric is given by
`U_(0)=(1)/(2)C_(0)V_(0)^(2)`
Note that here we have not used the expression
`U_(0)=(1)/(2)(Q_(0)^(2))/(C_(0))`
because here both charge and capacitance are changed whereas in equation 4 `V_(0)` remains constant .
After the dielectric is inserted the capacitance is increased hence the stored energy is also increased .
`U =(1)/(2) CV_(0)^(2) = (1)/(2)epsilon_(r)CV_(0)^(2)= epsilon_(r)U_(0)`
Since `e_(r) gt 1` we have `U gt U_(0)`
It may be noted here that since voltage between the capacitor `V_(0)` is constant the electric field between the plates also remains constant .