Question

(a) Distinguish with the help of a suitable diagram, the difference in the behaviour of a conductor and a dielectric placed in an external electric field. How does polarised dielectric modify the original external field ?
(b) A capacitor of capacitance C is charged fully by connecting it to a battery of emf E. If it is then disconnected from the battery. If the separation between the plates of the capacitor is now doubled, how will the following change ?
(i) Charge stored by the capacitor.
(ii) Field strength between the plates
(iii) Energy stored by the capacitor
Justify your answer in each case.

Answer 1

Hint: In order to solve the part (a) of given question first remember the properties of conductor and dielectric inside the electric field, which is following as –
When a conductor is placed in an external electric field then the total electric field inside the conductor becomes zero.
When a dielectric is placed in an external electric field then molecules of the dielectric become polarised. So, the total electric field of the dielectric is not equal to zero.
Using the above concept, we can draw the diagram.
For solving part (b) first use the law of conservation of charge for calculating the charge stored by the capacitor.
For calculating electric field we use formula of intensity of electric field due to conducting plates i.e., $E = \dfrac{\sigma }{{{ \in _0}}}$
For calculating energy stored in capacitor we use energy of parallel plate capacitor formula i.e., $U = \dfrac{{{Q^2}}}{{2C}}$
Where
C $ = $ capacitance of capacitor
Q $ = $ charge on plates of capacitor.

Complete step by step answer:
(a)

Conductor inside external electric field
(b)

Dielectric inside external electric field
From above diagram a and b shows the difference in behaviour of conductor and dielectric in an electric field.
In conductors there are free charge carriers. When an external electric field is applied on it then all free electrons move in the opposite direction of the external electric field and induce an electric field of the same strength in the opposite direction. So, the net electric field inside the conductor becomes zero.
In dielectric, there are no free charge carriers. A dipole moment is induced in the molecules of the dielectric on application of an external electric field. The induced opposing field is not strong enough to completely cancel the external electric field. Hence, a reduced value of electric field exists inside the material of dielectrics.
(b) Given a capacitor having capacity C, fully charged with a battery of emf E and charge Q at plates on the capacitor.
So, electric field inside the plates $E = \dfrac{\sigma }{{{ \in _0}}}$
Where
$\sigma = $ Surface charge density
${ \in _0} = $ Permittivity in vacuum and air
$\because \sigma = \dfrac{Q}{A}$
Where
Q $ = $ charge
A $ = $ cross-section area of plates
$\therefore E = \dfrac{Q}{{A{ \in _0}}}$ …..(1)
And energy stored in capacitor is
$V = \dfrac{1}{2}\dfrac{{{Q^2}}}{C}$
Where
C $ = $ capacitance of parallel plate capacitor and C is given as
$C = \dfrac{{A{ \in _0}}}{d}$
Where
d $ = $ separation between the plates
So, $U = \dfrac{{{Q^2}d}}{{2A{ \in _0}}}$
Now, according to the question , separation between the plates becomes doubled i.e., d becomes 2d then.
(i) Charge stored by the capacitor – Charge does not depend on separation between the plates because according to the law of conservation of charge, “for an isolated system, charge remains constant.”
So charge stored in the capacitor remains the same i.e., Q.
(ii) Field strength between the plates –
From equation 2
$E = \dfrac{Q}{{A{ \in _0}}}$
Hence separation becomes 2d but charge and area of cross section does not change. So, the electric field is also constant.
(iii) Energy stored in the capacitor
From equation 2
$U = \dfrac{{{Q^2}d}}{{2A{ \in _0}}}$
Here separation becomes 2d
So, energy will be
$ = \dfrac{{{Q^2}(2d)}}{{2A{ \in _0}}}$
$ = 2\left( {\dfrac{{{Q^2}d}}{{2A{ \in _0}}}} \right)$
$\therefore U' = 2U$

Hence, if separation is doubled then energy will also be doubled.

Note:
Here, one thing to be noted is that the capacitance of a capacitor depends on the area of plates and the ability of the dielectric to support electrostatic forces.
Larger plates provide greater capacity to store electric charge i.e., $C = \dfrac{{A{ \in _0}}}{d}$
As area increases, capacitance will be increases
And also inversely proportional to the distance between them.
$C \propto \dfrac{1}{d}$
As $d \uparrow C \downarrow $