Question

A parallel capacitor is charged by a battery. The battery is disconnected and a dielectric slab is inserted to completely fill the space between the plates. How will,
(a) Its capacitance
(b) Electric field between the plates and
(c) Energy stored in the capacitor is affected?
Justify your answer giving necessary mathematical expressions for each case.

Answer 1

Hint: In this question we have to find the effect of filling the space between the plates of the capacitor with dielectric slab and disconnecting the battery. In this question we will use the formula of capacitance of parallel plate capacitor.

Complete step by step answer:
If C is the capacitance of a parallel plate capacitor then the capacitance of a parallel plate capacitor is given by following formula,
$C = \dfrac{\varepsilon _0 A}{d}$
Where,
${\varepsilon _ 0}$ is the permittivity of free space,
d is the distance between the plates of the capacitor.
A is the area of the plates
Let after inserting a dielectric slab the capacitance becomes${C'}$, then the capacitance of the capacitor is given by following formula,
$\Rightarrow {C'} = \dfrac{{K{\varepsilon _ 0}A}}{d}$
Where,
K is the dielectric constant of the slab
So, from the above equation we see that the capacitance will increase K times of the initial.
If the battery is disconnected then the charge on the capacitor will remain constant. The charge on the capacitor is given by $Q = CV$. We have seen that the capacitance of the capacitor is increasing so the charge will also increase from this formula but the voltage will decrease.
As we know that the electric field is given by following formula,
$\Rightarrow E = \dfrac{V}{d}$
We have seen in the above explanation that the voltages are decreasing so according to the above formula the electric field will also decrease.
The energy stored in the capacitor is given by following formula,
$\Rightarrow U = \dfrac{{{Q^2}}}{{2C}}$
As we know that the charge will remain constant and capacitance is increasing, so the energy will decrease.
Result- The effect of disconnecting the battery and inserting a dielectric slab to completely fill the space between the plates has been explained above.

Note: In such types of questions if we have the knowledge of the formulae of charge, capacitance, electric field and energy stored; it becomes quite easy to find the effect of changing these parameters or changing the conditions of the capacitors. So we must remember the formulae.