Question

The electric field between the plates of a parallel plate capacitor when connected to a certain battery is ${E_0}$. If the space between the plates of the capacitor is filled by introducing a material of dielectric constant $K$ without disturbing the battery connections; the field between the plates will be:
A. $KE$
B. ${E_0}$
C. $\dfrac{{{E_0}}}{K}$
D. None of the above

Answer 1

Hint: An electric field is set-up between the plates due to laws of electrostatics which is given to be ${E_0}$ and when a dielectric slab is introduced another electric field is induced inside the slab which is lesser in magnitude and in the direction opposite to ${E_0}$. The net field is calculated by subtracting the electric field of the slab from ${E_0}$. The formula based on the concept of relative permittivity is applied to get the net electric field between the plates when there is a dielectric slab between them.

Complete step by step answer:
A parallel plate capacitor consists of two conducting plates separated by a distance that contain electrostatic charge which play a role of a capacitor by storing energy in the form of charge. When there is no dielectric slab in between the plates is an electric field that is set up due to the charges on the plates and this is already given to be ${E_0}$ in the question.The uniform electric field is set up only between the capacitor plates since the electric field at other points is zero. This formula is given by:
$E = \dfrac{\sigma }{{{\varepsilon _0}}}$
where, $E$ is the electric field, $\sigma $ is the surface charge density and ${\varepsilon _0}$ is the permittivity of free space which is a constant having the value $8.85 \times {10^{ - 12}}$.

When a dielectric slab of thickness $t$ which is lesser than the distance between the plates $d$ is inserted in between the parallel plates it undergoes polarization induced by the outer electric field${E_0}$ and hence there is an electric field set up inside the dielectric as well. This is denoted by, let’s say, ${E_p}$. The induced electric field due to polarization in the dipole is given by the formula:
${E_p} = \dfrac{{{\sigma _p}}}{{{\varepsilon _0}}}$
where, ${E_p}$ is the electric field due to polarization and ${\sigma _p}$ is the polarization density.

Polarization density is the induced dipole moment per unit volume when placed in an external electric field. The phenomenon of polarization induces negative charges to settle on the upper surface of the slab and the positive charges come towards the lower surface of the slab. Due to this effect, the electric field induced inside the slab is in the opposite direction with respect to the outer electric field ${E_0}$.
This difference in direction results in the net electric field to be:
${E_{net}} = {E_0} - {E_p}$ ---($1$)
This is because the strength and magnitude of the other electric field induced by the plates is greater than the field due to polarization in the slab. Hence, the net electric field will also be directed in the direction of ${E_0}$.

Now, the concept of relative permittivity or dielectric constant comes into picture. It is given by the ratio of the original field ${E_0}$ and the reduced field after inserting the dielectric slab which is ${E_{net}}$. Hence the equation is as follows:
$\kappa = \dfrac{{{E_0}}}{{{E_0} - {E_p}}}$ ----($2$)
where, $\kappa $ is the dielectric constant.
By taking the reciprocal the terms of equation ($2$) we get:
$\dfrac{1}{\kappa } = \dfrac{{{E_0} - {E_p}}}{{{E_0}}}$
By rearranging and cross multiplying the terms we get:
${E_0} - {E_p} = \dfrac{{{E_0}}}{\kappa }$ ----($3$)
Substituting equation ($3$) in ($1$) we get:
$\therefore {E_{net}} = \dfrac{{{E_0}}}{\kappa }$
Therefore, the net field inside the dielectric is $\dfrac{{{E_0}}}{\kappa }$.

The correct option is option C.

Additional information: A dielectric is a substance which does not allow the flow of charges through it which means that it is an insulator but however it permits the exertion of electrostatic forces on each other. It obeys the laws of induction of charges and is polarized through small displacement of charges. There are cases when the dielectric slab placed in between the plate has a thickness equivalent to the distance between the plates. This is a special case wherein the capacitance of the parallel plate capacitor increases.

Note:The direction of the polarized electric field opposes the electric field induced due to the plates and hence inserting a dielectric slab results in an overall reduction in the original field. The direction of the polarized electric field is in the direction of the outer electric field which is often neglected. The above equations are only for dielectric slabs that have a thickness lesser than the distance between the plates.