Question

On charging a parallel plate capacitor to a potential V, the spacing between the plates is Halved, and a dielectric medium ${ \in _1} = $ 10 is introduced between the plates, without disconnecting DC source. Explain, using suitable expression how the
1. Capacitance
2. Electric field and
3. Energy density of capacitor change

Answer 1

Hint: A parallel plate capacitor is charged to a potential V. After some time, spacing between the plates is halved and a dielectric medium of k=10 is introduced without disconnecting the DC source.

Complete Step-by-Step solution:
When the DC source remains connected, potential across the plates remains the same.
1. Initial capacitance, C=$\dfrac{{{ \in _0}A}}{d}$
When the spacing between the plate is halved d=d/2
A dielectric slab of K=10 is inserted.
New capacitance, C’=$k\dfrac{{{ \in _0}A}}{{d/2}}$=10
$
  \dfrac{{2{ \in _0}A}}{d} = 20C \\
    \\
$
New capacitance becomes 20 times that of the initial capacitance.
2. Electric field is given by E=$\dfrac{V}{d}$
When spacing is halved
New electric field becomes E’=$\dfrac{{2V}}{d}$=2E
Therefore, the electric field becomes twice that of the initial field.
3. Initial energy density=$\dfrac{1}{2}{ \in _0}{E^2}$
                               U=$\dfrac{1}{2}{ \in _0}{E^2}$
A dielectric slab of K=10 is inserted.
New energy density U’=$\dfrac{1}{2}K{ \in _0}{(2E)^2}$
                                         $ = \dfrac{1}{2}(10){ \in _0} \times 4{E^2}$
                                         =40$(\dfrac{1}{2}{ \in _0}{E^2})$
                                         =40U
Therefore, energy density becomes 40 times that of the initial energy density.

Note – Here in this problem you need to know the behaviour of the capacitor when it is connected to DC supply. So, when used in a direct current or DC circuit, a capacitor charges up to its supply voltage but blocks the flow of current through it because the dielectric of a capacitor is non-conductive and basically an insulator. At this point the capacitor is said to be “fully charged” with electrons.