Dielectric Constant

The dielectric constant, or relative permittivity, is a key concept in electromagnetism and materials science. It measures how effectively a material can store electrical energy within an electric field compared to a vacuum. This quantity is dimensionless and varies across different substances. A solid grasp of the dielectric constant is essential for applications such as capacitor design, insulation materials, and electronic devices. Materials with high dielectric constants are particularly valuable for improving the capacitance and performance of capacitors.

1.0Definition of Dielectric Constant $\left(\mathrm{K}\text{ or }{ϵ}_r\right)$

The dielectric constant, also termed as the Relative permittivity, quantifies how much a material can measure a material's capacity to store electrical energy in an electric field relative to a vacuum. It is a dimensionless quantity.

• The dielectric constant indicates how much a  material can enhance the capacitance of a capacitor by increasing its ability to store electrical energy, when it’s filled with a vacuum.
•  Materials with high dielectric constants can hold more electrical energy.

2.0Dielectric Constant in Terms of Permittivity

• It is defined as the ratio of the permittivity of material ($ϵ$) to the permittivity of free space or in Vacuum $\left({ϵ}_0\right)$
• ${ϵ}_r(K)=\frac{ϵ}{{ϵ}_0}=\frac{\text{ Permittivity of Material }}{\text{ Permittivity of Vacuum }}$
• Dielectric Constant of a medium decreases with increase in temperature, For Example Value of K for Water at $20^{\circ }\mathrm{C}=80$ and for water at $25^{\circ }C=78.5$
• Dielectric Constant is just a Number having No Units.
• Values of Dielectric Constant for Certain medium are Vacuum=1, Air=1.00059, Water=80, Metals=∞
• Dielectric Constant (K) is also named as Relative Permittivity $\left({ϵ}_r\right)$

3.0Dielectric Constant in Terms of Force

Coulomb Force in Vacuum 

$F_{\text{Vacuum }}=\frac{1}{4\pi {ϵ}_0}\frac{q_1{q}_2}{r^2}$ …(1)

Coulomb Force in Medium

$F_{\text{Medium }}=\frac{1}{4\pi ϵ}\frac{q_1{q}_2}{r^2}$ ….(2)

On dividing $\frac{F_{\text{Vacuum }}}{F_{\text{Medium }}}$ we get, $\frac{ϵ}{{ϵ}_0}={ϵ}_r(K)$

4.0Dielectric Constant in Terms of Electric Field Intensity

1. Value of Electric Field In the presence of Dielectric

• Two Parallel Conducting Plates equal and opposite charges separated by a certain distance having vacuum/air between the plates. If ($+\sigma$) surface charge density of positive plate and ($-\sigma$) surface charge density of Negative plate, Magnitude of Applied Electric field is given as $E_0=\frac{\sigma }{{ϵ}_0}$  …..(1)
• When a Non Polar dielectric slab inserted between the plates the phenomenon of Dielectric Polarisation takes place therefore the net effect is that the opposite faces of the dielectric will have equal and opposite charges, these charges are called induced charges and the process is called polarization of dielectric. Induced charges are bound charges and not the free charges.
• Induced charges set up an Electric Field which is known as Induced Electric Field

$\left(E_P\right)\overrightarrow{=}\frac{{\sigma }_P}{{ϵ}_0}$

• Net Electric Field inside the dielectric is given by,

$E=E_0-E_{P}E=\frac{\sigma }{{ϵ}_0}-\frac{{\sigma }_P}{{ϵ}_0}=\frac{\sigma -{\sigma }_P}{{ϵ}_0}$   ……..(2)

From equation (1) and (2) 

$K=\frac{E_0}{E}=\frac{\frac{\sigma }{{ϵ}_0}}{\frac{\sigma -{\sigma }_P}{E_0}}=\frac{\sigma }{\sigma -{\sigma }_P}$

$K=\frac{E_0}{E}=\frac{\sigma }{\sigma -{\sigma }_P}$

Induced Charge Density

${\sigma }_{\mathrm{p}}=\frac{-\mathrm{q}}{\mathrm{A}}\left(1-\frac{1}{\mathrm{K}}\right)$

$E=\frac{\sigma }{{\epsilon }_0}-\frac{q\left(1-\frac{1}{K}\right)}{A\times {\epsilon }_0}$

$\mathrm{E}=\frac{\sigma }{{\epsilon }_0}\left[1-1+\frac{1}{\mathrm{K}}\right]=\frac{\sigma }{{\epsilon }_0\mathrm{K}}=\frac{{\mathrm{E}}_0}{\mathrm{K}}$

• On placing a dielectric slab between two parallel plates, the electric field intensity inside the parallel plate gets reduced, the net electric field inside the dielectric is called the reduced value of electric field.

Dielectric Constant- It is  the ratio of applied electric field intensity \left(E_0\right) to the reduced value of electric field intensity (E) on placing a dielectric between two oppositely charged plates.

• $E_0>E$  The Value of K is always more than 1.
• Dielectric Constant is also known as Specific Inductive Capacity.

5.0Dielectric Constant in Terms of Capacitance

Capacitance of Parallel plate Capacitor having vacuum or air between the plates of the capacitor is given as 

$C=\frac{{ϵ}_{0}A}{d}$

On introducing Dielectric Slab of thickness ‘t’ between the plates of the capacitor

Voltage across the two plates of the capacitor is given by,

$V=E_0(d-t)+Et$

$V=E_0(d-t)+\frac{E_0}{K}t\left(\therefore E=\frac{E_0}{K}\right)$

$V=E_0\left(d-t+\frac{t}{K}\right)$

$V=\frac{q}{{ϵ}_{0}A}\left[(d-t)+\frac{t}{K}\right]\left(\therefore E_0=\frac{\sigma }{{ϵ}_0}=\frac{q}{{ϵ}_{0}A}\right)\left(\therefore \sigma =\frac{q}{A}\right)$

Capacitance of the Capacitor,

$C=\frac{q}{V}=\frac{q}{\frac{q}{{ϵ}_{0}A}\left[(d-t)+\frac{t}{K}\right]}=\frac{{ϵ}_{0}A}{d-t+\frac{t}{K}}$

$C=\frac{{ϵ}_{0}A}{d-t+\frac{t}{K}}$

If t=d, dielectric Slab fits in completely between the plates of the capacitor then capacitance is given by,

$C=\frac{{ϵ}_{0}A}{d-d+\frac{d}{K}}=K\frac{{ϵ}_{0}A}{d}$

$C=K\frac{{ϵ}_{0}A}{d}$

$C=K{C}_0\Rightarrow K=\frac{C}{C_0}$

$K=\frac{C}{C_0}=\frac{\text{ Capacitance of parallel plate capacitor with dielectric }}{\text{ Capacitance of parallel plate capacitor with vacuum or air }}$

Dielectric constant -It is defined as the ratio of the capacitance of a parallel plate capacitor with a dielectric slab between the plates to the capacitance of the same capacitor with a vacuum or air between the plates.

6.0Sample Questions on Dielectric Constant

Q-1. The dielectric constant of water is 80.Find its permittivity?

Solution:

$ϵ={ϵ}_0{ϵ}_r$

$ϵ=8.854\times 10^{-12}\times 80=7.08\times 10^{-10}{\mathrm{C}}^2{N}^{-1}{\mathrm{m}}^{-2}$

Q-2. The Coulomb’s Forces between two charges placed in water is F. Will the force between these two charges be increased or decreased or remain constant when the temperature of water increases? Explain?

Solution: Coulomb forces between two charges  separated by a separation r in a medium of relative permittivity K is given by, $F=\frac{1}{K}\frac{q_1{q}_2}{r^2}$

When temperature of medium (Water) increases, the value of dielectric constant decreases. Hence the value of F increases.

Q-3. A slab of material of relative permittivity K has the similar area as the plates of a parallel plate capacitor but has a thickness of 3d/4 where d is the separation of the plates. How is the capacitance changed when the slab is inserted between the plates?

Solution: Thickness(t)= $\frac{3d}{4}$

$C=\frac{C_0}{1-\frac{t}{d}\left(1-\frac{1}{K}\right)}=\frac{C_0}{1-\frac{\frac{3d}{4}}{d}\left(1-\frac{1}{K}\right)}=\frac{C_0}{1-\frac{3}{4}+\frac{3}{4K}}=\frac{C_0}{\frac{1}{4}+\frac{3}{4K}}=\frac{4K{C}_0}{K+3}$

where ${\mathrm{C}}_0=\frac{{\epsilon }_0\mathrm{A}}{\mathrm{d}}$

Q-4. Capacitance of a capacitor becomes $\frac{7}{6}$ times the original value, if a dielectric slab of thickness $t=\frac{2}{3}d$ is introduced between the plates. If d is the separation between the plates. Find the dielectric constant?

Solution:

$C=\frac{{ϵ}_{0}A}{d-t+\frac{t}{K}}$

$C^{\prime }=\frac{{ϵ}_{0}A}{d-\frac{2}{3}d+\frac{2d}{3K}}$

$\frac{7}{6}C=\frac{{ϵ}_{0}A}{d\left(1-\frac{2}{3}+\frac{2}{3K}\right)}=\frac{C}{\frac{K+2}{3K}}$

$\frac{7}{6}=\frac{3K}{K+2}$

7 K+14=18 K

$K=\frac{14}{11}$