Effect of Dielectric on Capacitance 

1.0Introduction

In physics, capacitors are devices that store electric charge and energy. The amount of charge a capacitor can store depends on its capacitance. One of the most important factors influencing capacitance is the dielectric material placed between the capacitor plates.

Understanding the effect of dielectric on capacitance is crucial for analyzing how capacitors function in circuits and for practical applications in electronics, power systems, and communication devices.

2.0Understanding Capacitance

Capacitance (C) is a measure of a capacitor's ability to store electric charge. It is defined as the ratio of the charge (Q) stored on its plates to the potential difference (V) across them:

$C=\frac{Q}{V}$

The SI unit of capacitance is the farad (F)

When a conductor is charged its potential increases. For an isolated conductor (conductor should be of finite dimension, so that potential of infinity can be assumed to be zero) potential of the conductor is proportional to charge given to it.

q = charge on conductor

V = potential of conductor

$q\propto V\Rightarrow q=CV$

3.0Formula for Capacitance of a Parallel Plate Capacitor

It consists of two large plane parallel conducting plates separated by a small distance.

Surface charge density σ = Q / A

Electric field intensity between plates, E₀ = σ / (2ε₀) + σ / (2ε₀) = σ / ε₀ = Q / (ε₀ A)

Potential difference between the plates, V = E₀ × d = Qd / (ε₀ A)

Capacitance of parallel plate capacitor, $C_0=\frac{Q}{V}=\frac{Q}{\left(\frac{Qd}{{\epsilon }_{0}A}\right)}=\frac{{\epsilon }_{0}A}{d}$

For a parallel plate capacitor without any dielectric material (in vacuum or air): C₀ = ε₀ A / d

where:

• ε₀ is the permittivity of free space $(8.85\times 10^{-12}F/m)$
• A is the area of the capacitor plates
• d is the distance between the plates

This formula shows that capacitance depends on the physical dimensions and the material between the plates (represented by ε₀​).

4.0What is a Dielectric?

A dielectric is an electrical insulator that can be polarized by an applied electric field. Unlike conductors, dielectrics do not have free electrons that can move throughout the material. Instead, the electrons are tightly bound to their atoms or molecules. When an external electric field is applied, the positive and negative charges within the dielectric molecules are slightly displaced, creating microscopic electric dipoles.

Common dielectric materials include glass, mica, paper, plastics, and ceramics. They are crucial for both preventing the plates from short-circuiting and for increasing the capacitance of the device.

The Role of a Dielectric in a Capacitor

When a dielectric material is inserted between the plates of a capacitor, it does three primary things:

1. It physically separates the plates, preventing them from touching and creating a short circuit.
2. It provides mechanical support for the capacitor structure.
3. It significantly increases the capacitance of the capacitor. This is the most important electrical effect.

5.0Effect of Dielectrics

• The insulators in which at microscopic level displacement of charges takes place in presence of electric field are known as dielectrics.
• Dielectrics are non-conductors up to certain value of field depending on its nature. If the field exceeds this limiting value called dielectric strength they lose their insulating property and begin to conduct.
• Dielectric strength is defined as the maximum value of electric field that a dielectric can tolerate without breakdown. Unit is volt/metre. Dimensions M¹ L¹ T⁻³ A⁻¹.

6.0Polar Dielectrics

• In absence of external field, the centres of positive and negative charge do not coincide due to asymmetric shape of molecules.
• Each molecule has permanent dipole moment.
• The dipoles are randomly oriented so average dipole moment per unit volume of polar dielectric in absence of external field is nearly zero.
• In presence of external field dipoles tend to align in direction of field.
• Example: Water, Alcohol, CO₂, HCl, NH₃

7.0Nonpolar Dielectrics

8.0Polarisation

The alignment of dipole moments of permanent or induced dipoles in the direction of applied electric field is called polarisation.

Polarisation Vector ($\vec{P}$)

This is a vector quantity which describes the extent to which molecules of dielectric become polarized by an electric field or oriented in direction of field.

$\vec{P}=\text{the dipole moment per unit volume of dielectric}=n\vec{p}$

where n is number of atoms per unit volume of dielectric and p is dipole moment of an atom or molecule.

$\vec{P}=n\vec{p}=\frac{q_{i}d}{Ad}=\left(\frac{q_i}{A}\right)={\sigma }_i=\text{induced surface charge density.}$

Effect on Capacitance: The Dielectric Constant

When a dielectric is introduced between the plates of a capacitor, the capacitance changes depending on whether the dielectric partially or completely fills the space.

(a) Dielectric Completely Filling the Space

Electric field intensity in vacuum ⇒ E₀
Electric field intensity in medium ⇒ Eₘ

Eₘ = σ / ε = σ / (ε₀ εᵣ) = Q / (ε₀ εᵣ A) ⇒ Eₘ = E₀ / εᵣ

Potential difference between the plates V = Eₘ × d = Qd / (ε₀ εᵣ A)

Capacitance of a capacitor in presence of medium Cₘ = Q / V = (ε₀ εᵣ A (Q)) /Q.d = εᵣ C₀

If the entire space between plates is filled with a dielectric of dielectric constant K:

C = K C₀ = (K ε₀ A )/ d

Here:

• K = Dielectric constant (Relative Permittivity, always >1)
• C₀ = Capacitance without dielectric

Thus, capacitance increases K times compared to air or vacuum.

(b) Dielectric Partially Introduced

Surface charge density, σ = Q / A

Electric field in air or vacuum, E₀ = σ / ε₀ = Q / (ε₀ A)

Electric field in dielectric medium, Eₘ = σ / ε = Q / (ε₀ εᵣ A)

Potential difference between the plates of capacitor:

$V=E_0(d-t)+E_{m}t=\frac{Q}{{\epsilon }_{0}A}(d-t)+\frac{Q}{{\epsilon }_0{\epsilon }_{r}A}t=\frac{Q}{{\epsilon }_{0}A}\left[(d-t)+\frac{t}{{\epsilon }_r}\right]$

Capacitance, C = Q / V

$\Rightarrow C=\frac{Q}{\frac{Q}{{\epsilon }_{0}A}\left[(d-t)+\frac{t}{{\epsilon }_r}\right]}=\frac{{\epsilon }_{0}A}{(d-t)+\frac{t}{{\epsilon }_r}}$

In case of multiple slabs,

$C=\frac{{\epsilon }_{0}A}{\left(d-(t_1+t_2+\dots )\right)+\frac{t_1}{{\epsilon }_r}+\frac{t_2}{{\epsilon }_r}+\dots }$

If the dielectric slab of thickness t is introduced (where t<d):

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

This shows how capacitance varies with the thickness and dielectric constant of the material.

(c) Important Observations

• Insertion of dielectric increases capacitance.
• The higher the dielectric constant K, the greater the capacitance.
• Air or vacuum acts as a dielectric with K=1.

Mechanisms Behind the Effect of Dielectrics

The increase in capacitance is a consequence of the dielectric's interaction with the electric field between the capacitor plates.

Polarization of Dielectric Material

When a dielectric is placed in an external electric field ($E_0$​), the molecules within it become polarized. This means that the positive ends of the molecules align with the external field, and the negative ends align against it. This alignment creates a new, internal electric field ($E_{induced}$) within the dielectric that opposes the original external field.

Reduction of Electric Field

Because the induced electric field $(E_{induced})$ opposes the external electric field ($E_0$), the net electric field (E) inside the capacitor is reduced.

$E=E_0-E_{induced}=\frac{E_0}{\chi }$

The net electric field inside the dielectric is only a fraction (1/$\chi$) of the original field without the dielectric.

Effect on Voltage

Since the potential difference (V) between the plates is related to the electric field (E) by V=E×d (for a uniform field), the reduction in the electric field leads to a reduction in the potential difference across the plates.

$V=E\times d=\frac{E_0}{\chi }\times d=\frac{V_0}{\chi }$

The new potential difference (V) is reduced by the same factor, κ, as the electric field. This reduction in voltage is the key to the increase in capacitance.

9.0Relation between Polarisation and Surface Charge Density of Induced Charge

Equivalent dipole moment of dielectric slab = qᵢ × d

Electric polarization

$(\vec{P})=\frac{q_i\times d}{V}(V\Rightarrow \text{volume of dielectric slab})$

$\vec{P}=\frac{q_i\times d}{A\times d}\Rightarrow \vec{P}=\frac{q_i}{A}={\sigma }_i$

P = σᵢ

It is clear that polarisation is equal to induced surface charge density.

Electric susceptibility (χₑ)

Polarisation (P) of dielectrics is directly proportional to the electric field (E), i.e.,

P ∝ E

P = ε₀ χₑ E

Where, χₑ ⇒ electric susceptibility (constant)

For vacuum, χₑ = 0

Dielectric strength

The maximum value of the electric field intensity that can be applied to the dielectric without its electric breakdown.

$E_{\text{break}}=\frac{V_{\text{break}}}{d}$

$\text{For air, }{E}_{\text{break}}=3\times 10^{6}V/m$

Effects of Dielectrics in Capacitor

Distance and Area Division by Dielectric

(1) Distance Division

(a) Distance is divided and area remains same

(b) Capacitors are in series

(c) Individual capacitances are

C₁ = ε₀ εᵣ₁ A / d₁
C₂ = ε₀ εᵣ₂ A / d₂

These two are in series

$\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}\Rightarrow \frac{1}{C}=\frac{d_1}{{\epsilon }_0{\epsilon }_{r_1}A}+\frac{d_2}{{\epsilon }_0{\epsilon }_{r_2}A}$

$\Rightarrow \frac{1}{C}=\frac{1}{{\epsilon }_{0}A}\left[\frac{d_1{\epsilon }_{r_2}+d_2{\epsilon }_{r_1}}{{\epsilon }_{r_1}{\epsilon }_{r_2}}\right]$

$\Rightarrow C={\epsilon }_{0}A\left[\frac{{\epsilon }_{r_1}{\epsilon }_{r_2}}{d_1{\epsilon }_{r_2}+d_2{\epsilon }_{r_1}}\right]$

Special case: If $d_1=d_2=\frac{d}{2}$

$C=\frac{{\epsilon }_{0}A}{d}\left[\frac{2{\epsilon }_{r_1}{\epsilon }_{r_2}}{{\epsilon }_{r_1}+{\epsilon }_{r_2}}\right]$

(2) Area Division

(a) Area is divided and distance remains same.

(b) Capacitors are in parallel.

(c) Individual capacitances are

C₁ = ε₀ εᵣ₁ A₁ / d
C₂ = ε₀ εᵣ₂ A₂ / d

These two are in parallel so, C = C₁ + C₂

$C=\frac{{\epsilon }_0{\epsilon }_{r_1}{A}_1}{d}+\frac{{\epsilon }_0{\epsilon }_{r_2}{A}_2}{d}$

Special case: If A₁ = A₂ = A / 2

$\text{Then,}C=\frac{{\epsilon }_{0}A}{d}\left(\frac{{\epsilon }_{r_1}+{\epsilon }_{r_2}}{2}\right)$

10.0Comparison of  (Electric Field),  (Surface Charges Density) and Other Parameters

Comparison of E (electric field), σ (surface charge density), Q (charge), C (capacitance) before and after inserting a dielectric slab between the plates of a parallel plate capacitor

Case 1: When constant potential difference across capacitor plates is maintained.

Parallel plate capacitor of capacitance C connected to a voltage source V due to which the capacitor is charged to the steady state charge q = CV. As battery remains connected during insertion of slab the potential difference across capacitor does not change in this case.

Initial state:

(a) Initial Capacitance, C = ε₀ A / d

(b) Initial charge, Q = CV

(c) Initial electric field, E = σ / ε₀ = CV / (A ε₀)

(d) Initially potential difference between the plates

Ed = V ⇒ E = V / d

⇒ V / d = σ x V/ d x ε₀ ⇒ σ / ε₀ = σ′ / (K ε₀)

⇒ σ′ = Kσ

(e) Energy stored in capacitor = CV² / 2

Final state:

(a) Final capacitance, C′ = A ε₀ K / d

(b) Q′ = C′V

(c) E′ = σ′ / (K ε₀) = CV / (A ε₀)

(d) Here potential difference between the plates: E′d = V ⇒ E′ = V / d

(e) Energy stored in capacitor = KCV² / 2

Or we can say that after inserting dielectric final charge becomes KCV which is K times the initial charge. Hence CV(K − 1) charge has flown through battery and work done by battery is CV²(K − 1) in the insertion process.

Energy equation for the capacitor circuit in dielectric insertion process will be$W_{batt}+W_{ext}=\Delta U+Heat$

If dielectric is inserted slowly and circuit has negligible resistance then heat loss will be negligible and some negative work will be done by external force.

Case 2: When plates of capacitor have constant charge:

Parameter before insertion of dielectric:

(a) Initial capacitance C = ε₀ A / d

(b) Initial charge = Q

(c) Initial electric field = E

(d) Initial voltage = V

(e) Initial energy stored = U

Parameter after insertion of dielectric:

(a) Final capacitance = kC

(b) Final charge = Q

(c) Final electric field = E / k

(d) Final voltage = V / k

(e) Final energy stored = U / k

11.0Applications of Dielectrics in Capacitors

1. Energy Storage: Used in capacitor banks for power supply systems.
2. Oscillatory Circuits: Radio tuners and communication systems use capacitors with dielectrics for frequency control.
3. Insulating Materials: Dielectrics act as insulators, preventing direct conduction between capacitor plates.
4. Capacitor Design: High K dielectrics (like ceramic) are used in compact capacitors for higher efficiency.
5. Medical Equipment: Capacitors with dielectrics are used in defibrillators to store and release large amounts of energy quickly.

12.0Sample Questions on Effect of Dielectric on Capacitance

Question 1: Find out capacitance between A and B if three dielectric slabs of dielectric constant K₁ of area A₁ and thickness d, K₂ of area A₂ and thickness d₁ and K₃ of area A₂ and thickness d₂ are inserted between the plates of parallel plate capacitor of plate area A as shown in figure. (Given distance between the two plates d = d₁ + d₂).

Solution:

It is equivalent to $C=C_1+\frac{C_2{C}_3}{C_2+C_3}$

$C=\frac{A_1{K}_1{\epsilon }_0}{d_1+d_2}+\frac{\left(\frac{A_2{K}_2{\epsilon }_0}{d_1}\right)\left(\frac{A_2{K}_3{\epsilon }_0}{d_2}\right)}{\left(\frac{A_2{K}_2{\epsilon }_0}{d_1}\right)+\left(\frac{A_2{K}_3{\epsilon }_0}{d_2}\right)}$

$=\frac{A_1{K}_1{\epsilon }_0}{d_1+d_2}+\frac{A_2^2{K}_2{K}_3{\epsilon }_0^2}{A_2{K}_2{\epsilon }_0{d}_2+A_2{K}_3{\epsilon }_0{d}_1}$

$=\frac{A_1{K}_1{\epsilon }_0}{d_1+d_2}+\frac{A_2{K}_2{K}_3{\epsilon }_0}{K_2{d}_2+K_3{d}_1}$

Question 2: A parallel plate isolated condenser consists of two metal plates of area A and separation d. A slab of thickness t and dielectric constant K is inserted between the plates with its faces parallel to the plates and having the same surface area as that of the plates. Find the capacitance of the system. If K = 2, for what value of t/d will the capacitance of the system be 3/2 times that of the condenser with air filling the full space? Calculate the ratio of the energy in the two cases and account for the energy change (assuming charge on the plate to be constant).

Solution:

$C^{\prime }=\frac{A{\epsilon }_0}{\frac{t}{K}+d-t}$

(i) Without dielectric, C = Aε₀ / d

With dielectric, C′ = Aε₀ / (t/2 + d − t) = 3C / 2 = (3/2)(Aε₀ / d) ⇒ d / t = 3 / 2

(ii) Energy = q² / (2C)

Energy in 1st case, E₁ = q² / (2C)

Energy in 2nd case, E₂ = q² / (2C′)

$\frac{E_1}{E_2}=\frac{C^{\prime }}{C}=\frac{3}{2}$

(iii) $\Delta E=E_2-E_1=\frac{q^2}{2C^{\prime }}-\frac{q^2}{2C}=\frac{q^2}{2}\left[\frac{2}{3C}-\frac{1}{C}\right]=-\frac{q^2}{6C}=-\frac{q^{2}d}{6A{\epsilon }_0}$

Question 3: Find out capacitance between A and B if two dielectric slabs of dielectric constant K₁ and K₂ of thickness d₁ and d₂ and each of area A, are inserted between the plates of parallel plate capacitor of plate area A as shown in figure.

Solution:

C = σA / V (C = Q / V ; q = σA)

$V=E_1{d}_1+E_2{d}_2=\frac{\sigma d_1}{K_1{\epsilon }_0}+\frac{\sigma d_2}{K_2{\epsilon }_0}=\frac{\sigma }{{\epsilon }_0}\left(\frac{d_1}{K_1}+\frac{d_2}{K_2}\right)$

$C_1=\frac{K_1{\epsilon }_{0}A}{d_1}{C}_2=\frac{K_2{\epsilon }_{0}A}{d_2}$

$\therefore C=\frac{A{\epsilon }_0}{\frac{d_1}{K_1}+\frac{d_2}{K_2}}$

$\Rightarrow \frac{1}{C}=\frac{d_1}{A{K}_1{\epsilon }_0}+\frac{d_2}{A{K}_2{\epsilon }_0}$

This formula suggests that the system between A and B can be considered as a series combination of two capacitors.

Question 4: Two parallel plate air capacitors each of capacitance C were connected in series to a battery with e.m.f. ε. Then one of the capacitors was filled up with uniform dielectric of relative permittivity k. How many times did the electric field strength in that capacitor decrease? What amount of charge flows through the battery?

Solution:

$V_1=\frac{Q_1}{kC}=\frac{\epsilon }{k+1},E_1=\frac{V_1}{d}=\frac{1}{k+1}\cdot \frac{\epsilon }{d}\text{and}E=\frac{Q}{dC}=\frac{\epsilon }{2d}$

$E_1=\frac{2}{(k+1)}E\Rightarrow \text{2(1+k) times decrease}$

$\Delta q=Q_1-Q=\frac{kC\epsilon }{k+1}-\frac{C\epsilon }{2}=\frac{C\epsilon (2k-k-1)}{2(k+1)}$

$\Rightarrow \Delta q=\frac{1}{2}\frac{C\epsilon (k-1)}{(k+1)}$