Effect of Dielectric on Capacitance: Formula, Explanation, and Examples

18.8 DIELECTRIC

18.8.1 Dielectric

1. A dielectric is an insulating material. Hence no free electrons are available in it.
2. Very often a kind of dielectric (besides air) is placed between the plates of a parallel plate capacitor.
3. Examples of some dielectrics used in a capacitor: air, waxed paper, plastics, mica, ceramic.

Most capacitors have a nonconducting material, or dielectric, between their conducting plates. A common type of capacitor uses long strips of metal foil for the plates, separated by strips of plastic sheet such as Mylar. A sandwich of these materials is rolled up, forming a unit that can provide a capacitance of several microfarads in a compact package (Fig. 1). 

Figure.18.30: A common type of capacitor uses dielectric sheets to separate the conductors.

18.8.2 Non-polar and Polar Dielectrics

1. Non-polar dielectric

According to the classical theory of atoms, electrons revolve round the nucleus of a neutral atom. For an atom of a non-polar dielectric, the centre of the negative charge of the electrons 'coincides'  with the centre of the positive charge of the nucleus (Figure 18.31(a)). It does not become a permanent dipole.

Figure 18.31

2. Polar dielectric

(a) Consider the molecule of water. Its two positively charged hydrogen ions are attached to a negatively charged oxygen ion (Figure 18.31(b)). Such an arrangement of ions causes the centre of the negative charge to be permanently separated slightly away from the centre of the positive charge, thus forming a permanent dipole.

(b) At normal temperature and in the absence of electric field, these dipoles are randomly aligned (Figure 18.31(c)).

18.8 Dielectric in Capacitors

18.8.3 Electric Polarisation

Non-polar dielectric

Suppose an electric field is applied to a non-polar dielectric. The electric field will cause the centre of the negative charge of the atom to be displaced slightly away from the centre of the positive charge, thus producing a temporary dipole. Furthermore, these dipoles align themselves along the direction of the electric field, as shown in Figure 18.32. The separation of two opposite charges slightly away from each other due to the application of an electric field is known as electric polarisation.

Figure 18.32 Electric polarisation in a non-polar dielectric: formation and alignment of dipoles under an applied electric field.

Polar dielectric

When an electric field is applied to a polar dielectric, the permanent dipoles in the substance align themselves along the direction of the electric field. Without the field the dipoles do not align themselves towards a specific direction.

18.8.4 Dielectric Breakdown

1. Dielectric breakdown

(a) Suppose an external electric field is applied to a block of dielectric. Let the strength of the field increase gradually. When the field strength becomes too high, a sudden surge of large current will flow through the dielectric. The dielectric is said to have experienced dielectric breakdown when this occurs.

(b) Dielectric breakdown can occur to the dielectric placed between the plates of a capacitor if the p.d. applied across the capacitor gets too high. Most likely the capacitor would be damaged permanently if ever dielectric breakdown occurs.

2. Dielectric strength

(a) The maximum electric field strength, or electric potential gradient, that a dielectric can withstand without producing dielectric breakdown is known as the dielectric strength.

(b) Dielectric strength is often expressed in V mm^-1.

3. Some typical values of dielectric strength

• Air : 0.3 kV mm^-1
• Ceramic : 39 kV mm^-1
• Paper : 49 kV mm^-1
• Mica : 59 kV mm^-1
• Glass : 79 kV mm^-1

18.8.5 Effect of Dielectric on Electric Field in Capacitor

A dc voltage is applied to an air-filled parallel plate capacitor so that one plate is charged positively while the other plate is charged negatively. A uniform electric field of strength E₀ is set up in the air space between the plates, as shown in Figure 18.32(a). The strength of the field will be affected if the air is replaced with a solid or liquid dielectric.

Reasons:

(a) Suppose the space between the plates is completely filled with a kind of dielectric PQRS, as shown in Figure 18.32(b). Since an electric field is present in the dielectric, electric polarisation occurs. Temporary dipoles will be set up if the dielectric is a non-polar one. The dipoles align themselves along the direction of the electric field.

(b) Due to the alignment of the dipoles, the surface PQ of the dielectric which is near to the positively charged plate of the capacitor becomes negatively charged. The surface RS which is near to the negatively charged plate becomes positively charged. However, the bulk of the dielectric is effectively neutral.

(c) The negative surface charge on PQ and the positive surface charge on RS can set up an electric field of strength E_p. The direction of this induced field is opposite to the direction of the applied field, as shown in Figure 18.32(c).

(d) The opposition of the induced field to the applied field will reduce the strengh of the field in the electric. Hence, the insertion of a dielectric in the air space between a parallel plate capacitor will reduce the strength of the electric field existing in the space.

18.8.6 Effect of Dielectric on p.d. and Capacitance

A d.c. voltage $V_0$ is applied across an air-filled parallel plate capacitor of capacitance $C_0$. The charge stored in the capacitor is Q. An electric field of strength E₀exists in the air space between the plates, as shown in Figure 18.30(a), and is given by

$E_0 = \frac{V_0}{d}$

The air space is then filled completely with a dielectric. The insertion of the dielectric brings about the following effects to the capacitor.

(a) p.d. across the charged capacitor (not connected to any voltage supply)

An induced electric field is set up between the plates. As a result, the resultant electric field of strength Eis weaker than the initial field of strength $E_0$. As shown in Figure 18.30(b), we can relate the two strengths through the relation

$E = \left(\frac{1}{\epsilon_r} \right) E_0$

where $\epsilon_r$ is known as the relative permittivity or dielectric constant of the dielectric, and $\epsilon_r > 1$. But we have

$E = \frac{V}{d}$

where V is the p.d. across the capacitor after the insertion of the dielectric.

$\frac{V}{d} = \left(\frac{1}{\epsilon_r} \right) \frac{V_0}{d}$

This implies that $V < V_0$

We have a charged air-filled parallel plate capacitor which is not connected to a d.c. battery. Upon insertion of a dielectric in the space between the plates, the p.d. across the capacitor will decrease.

(b) Capacitance of capacitor

The capacitance $C_0$ of a charged air-filled parallel plate capacitor is given by

$C_0 = \frac{Q}{V_0}$

where Q is the charge stored in the capacitor and $V_0$ is the p.d. across the capacitor.

Suppose that the capacitor is not connected to a battery or any other external circuit. Then the charge stored will not change, i.e.

$Q = constant$

Let us insert a dielectric into the space between the plates of the capacitor. After inserting the dielectric, the p.d. across the capacitor will decrease to a new value, V. The new capacitance C of the capacitor is given by

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

$Q = CV$

$CV = C_0 V_0$

But we have

$V < V_0$

This means that we must have

$C > C_0$

The capacitance is greater when there is a dielectric in the space between the plates.

The new capacitance C is given by

$C = \epsilon_r C_0$

where $\epsilon_r$ is the relative permittivity of the dielectric.

$C = \frac{\epsilon_r \epsilon_0 A}{d}$

Note: Air itself is a dielectric. For air, $\epsilon_r \cong 1.0$

The dielectric constant is a pure number. Because is always greater than is always greater than unity. Some representative values of are given in Table 1. For vacuum, by definition. For air at ordinary temperatures and pressures, is about 1.0006; this is so nearly equal to 1 that for most purposes an air capacitor is equivalent to one in vacuum. Note that while water has a very large value of it is usually not a very practical dielectric for use in capacitors. The reason is that while pure water is a very poor conductor, it is also an excellent ionic solvent. Any ions that are dissolved in the water will cause charge to flow between the capacitor plates, so the capacitor discharges.

Dielectric Constant Table and Types of Capacitors

Table 1: Here is the rewritten table of dielectric constant values K at 20°C:

Thus, we see that a dielectric provides the following advantages:

• Increase in capacitance
• Increase in maximum operating voltage
• Possible mechanical support between the plates, which allows the plates to be close together without touching, thereby decreasing d and increasing C.

Types of Capacitors

Commercial capacitors are often made from metallic foil interlaced with thin sheets of either paraffin-impregnated paper or Mylar as the dielectric material. These alternate layers of metallic foil and dielectric are rolled into a cylinder to form a small package (Fig. 4a). High-voltage capacitors commonly consist of a number of interwoven metallic plates immersed in silicone oil (Fig. 4b). Small capacitors are often constructed from ceramic materials.

Fig. 4: Three commercial capacitor designs. (a) A tubular capacitor, whose plates are separated by paper and then rolled into a cylinder. (b) A high-voltage capacitor consisting of many parallel plates separated by insulating oil. (c) An electrolytic capacitor.

Often, an electrolytic capacitor is used to store large amounts of charge at relatively low voltages. This device, shown in Figure 4c, consists of a metallic foil in contact with an electrolyte—a solution that conducts electricity by virtue of the motion of ions contained in the solution. When a voltage is applied between the foil and the electrolyte, a thin layer of metal oxide (an insulator) is formed on the foil, and this layer serves as the dielectric. Very large values of capacitance can be obtained in an electrolytic capacitor because the dielectric layer is very thin, and thus the plate separation is very small.

Electrolytic capacitors are not reversible as are many other capacitors—they have a polarity, which is indicated by positive and negative signs marked on the device. When electrolytic capacitors are used in circuits, the polarity must be aligned properly. If the polarity of the applied voltage is opposite that which is intended, the oxide layer is removed and the capacitor conducts electricity instead of storing charge.

Fig. 5: A variable capacitor. When one set of metal plates is rotated so as to lie between a fixed set of plates, the capacitance of the device changes.

Variable capacitors (typically 10 to 500 pF) usually consist of two interwoven sets of metallic plates, one fixed and the other movable, and contain air as the dielectric (Fig. 5). These types of capacitors are often used in radio tuning circuits.

Example Problems: Effect of Dielectric on Capacitance

Example 18.13

The space between the plates of an air-filled parallel plate capacitor is completely filled with mica. Determine the percentage change in capacitance of the capacitor due to the insertion of mica into the space mentioned. (For mica, ε_r = 5)

Answer

Air-filled capacitor:

$C_0 = \frac{\epsilon_0A}{d}$

Mica-filled capacitor:

$C_1 = \frac{\epsilon_r \epsilon_0A}{d}$

fractional change in capacitance

$=\frac{C_1 − C_0}{C_0}$

$= \frac{C_1 }{C_0}− 1$

$=\epsilon_r − 1$

$= 5 − 1 = +4$

percentage increase in capacitance = +400%

Example 18.14

A parallel-plate capacitor has capacitance C₀ = 5.00 pF when there is air between the plates. The separation between the plates is 1.50 mm. (a) What is the maximum magnitude of charge Q that can be placed on each plate if the electric field in the region between the plates is not to exceed 3.00 × 10⁴ V/m? (b) A dielectric with K = 2.70 is inserted between the plates of the capacitor, completely filling the volume between the plates. Now what is the maximum magnitude of charge on each plate if the electric field between the plates is not to exceed 3.00 × 10⁴ V/m?

Answer:

Given:

Capacitance without the dielectric: C₀ = 5.00 pF = 5.00 × 10⁻¹² F

Separation between the plates: d = 1.50 mm = 1.50 × 10⁻³ m

Maximum electric field: E_max = 3.00 × 10⁴ V/m

(a) Maximum charge without the dielectric

We use the following relations:

The relationship between electric field and voltage:

V_max = E_max · d

where V_max is the maximum voltage, E_max is the maximum electric field, and d is the separation between the plates.

Substituting the values:

V_max = (3.00 × 10⁴ V/m)(1.50 × 10⁻³ m)

V_max = 45.0 V

Now, we use the relation between charge, capacitance, and voltage:

Q = C₀ · V_max

Substituting the values:

Q = (5.00 × 10⁻¹² F)(45.0 V)

Q = 2.25 × 10⁻¹⁰ C

So, the maximum charge that can be placed on each plate without exceeding the electric field is: Q = 2.25 × 10⁻¹⁰ C.

(b) Maximum charge with the dielectric

When a dielectric with K = 2.70 is inserted, the capacitance increases by a factor of K, and the maximum charge will also increase.

The new capacitance with the dielectric is:

C = K · C₀

C = (2.70)(5.00 × 10⁻¹² F) = 1.35 × 10⁻¹¹ F

The electric field in the dielectric is still E_max, but now the voltage will be different due to the increased capacitance. We can calculate the new voltage with the dielectric:

V_max = E_max · d

(Note that this voltage is the same as in part (a) because the field and the separation are the same.)

V_max = 45.0 V

Finally, the maximum charge on each plate with the dielectric is:

Q = C · V_max

Substituting the values:

Q = (1.35 × 10⁻¹¹ F)(45.0 V)

Q = 6.08 × 10⁻¹⁰ C

So, the maximum charge on each plate with the dielectric inserted is: Q = 6.08 × 10⁻¹⁰ C

Example 18.15

Two parallel plates have equal and opposite charges. When the space between the plates is evacuated, the electric field is $E =3.20 \times 10^5 \ V/m$. When the space is filled with dielectric, the electric field is $E =2.50 \times 10^5 \ V/m$. (a) What is the charge density on each surface of the dielectric? (b) What is the dielectric constant?

Solution:

Given:

Electric field in the vacuum: $E_0=3.20 \times 10^5 \ V/m$

Electric field with dielectric: $E=2.50 \times 10^5 \ V/m$

(a) Charge density on each surface of the dielectric

When the space is evacuated, the electric field is related to the charge density on the plates. The electric field between two parallel plates with charge density $\sigma$ is given by the relation:

$E_0=\frac{\sigma}{\epsilon_0}$

From this, we can solve for $\sigma$ (the charge density):

$\sigma=\epsilon_0.E_0$

Substituting the given values:

$\sigma=(8.85 \times 10^{-12} \ F/m)(3.20 \times 10^5 \ V/m)$

$\sigma=2.832 \times 10^{-6} \ C/m^2$

So, the charge density on the plates is $2.832 \times 10^{-6} \ C/m^2$

Now, when the dielectric is inserted, the charge density on the surface of the dielectric is the same as the charge density on the plates. However, the electric field in the dielectric is reduced due to the dielectric constant K. The relation between the electric field in the vacuum $E_0$ and the electric field in the dielectric $E$ is:

$E=\frac{E_0}{K} \rightarrow K=\frac{E_0}{E}$

Substitute the given values:

$K=\frac{3.20 \times 10^5 \ V/m}{2.50 \times 10^5 \ V/m}=1.28$

Thus, the dielectric constant K is 1.28.

Example 18.16

Suppose the parallel plates in Fig. 3 each have an area of $2000 \ cm^2$ ($2.00 \times 10^{-1} \ m^2$) and are 1.00 cm ($1.00 \times 10^{-2} \ m$) apart. We connect the capacitor to a power supply, charge it to a potential difference $V_0=3.00 \ kV$ and disconnect the power supply. We then insert a sheet of insulating plastic material between the plates, completely filling the space between them. We find that the potential difference decreases to 1.00 kV while the charge on each capacitor plate remains constant. Find (a) the original capacitance $C_0$, (b) the magnitude of charge $Q$ on each plate; (c) the capacitance $C$ after the dielectric is inserted; (d) the dielectric constant $K$ of the dielectric; (e) the permittivity $\epsilon_0$ of the dielectric; (f ) the magnitude of the induced charge $Q_i$ on each face of the dielectric; (g) the original electric $E_0$ field between the plates; and (h) the electric field $E$ after the dielectric is inserted.

Answer

(a) With vacuum between the plates, we use Eq. $C_0=\frac{\epsilon_0 A}{d}$ with $K=1$:

$C_0=\frac{\epsilon_0 A}{d}=\frac{(8.85 \times 10^{-12} \ F/m)(2.00 \times 10^{-1} \ m^2)}{(1.00 \times 10^{-2} \ m)}$

$C_0=1.77 \times 10^{-10} \ F=177 \ pF$

(b) From the definition of capacitance:

$Q=C_0V_0=(1.77 \times 10^{-10} \ F)(3.00 \times 10^3 \ V)$

$Q=5.31 \times 10^{-7} \ C=0.531 \ \mu C$

(c) When the dielectric is inserted, Q is unchanged but the potential difference decreases to V =1.00 kV:

$C=\frac{Q}{V}=\frac{5.31 \times 10^{-7} \ C}{1.00 \times 10^3 \ V}$

$C=5.31 \times 10^{-10} \ F=531 \ pF$

(d) Dielectric constant:

$K=\frac{C}{C_0}=\frac{5.31 \times 10^{-10}}{1.77 \times 10^{-10}}=3.00$

Alternatively:

$K=\frac{V_0}{V}=\frac{3000}{1000}=3.00$

(e) Permittivity:

$\epsilon=K\epsilon_0=(3.00)(8.85 \times 10^{-12})$

$\epsilon=2.66 \times 10^{-11} \ C^2/N.m^2$

(f) Induced charge:

$Q_i=Q\left(1-\frac{1}{K}\right)=(5.31 \times 10^{-7})\left(1-\frac{1}{3.00}\right)$

$Q_i=3.54 \times 10^{-7} \ C$

(g) Original electric field:

$E_0=\frac{V_0}{d}=\frac{3000}{1.00 \times 10^{-2}}=3.00 \times 10^5 \ V/m$

(h) Electric field after dielectric:

$E=\frac{V}{d}=\frac{1000}{1.00 \times 10^{-2}}=1.00 \times 10^5 \ V/m$

or,

$E=\frac{\sigma}{\epsilon}=\frac{Q}{\epsilon A}$

$E=\frac{(5.31 \times 10^{-7})}{(2.66 \times 10^{-11})(2.00 \times 10^{-1})}$

$E=1.00 \times 10^5 \ V/m$

or,

$E=\frac{\sigma - \sigma_i}{\epsilon_0}=\frac{Q-Q_i}{\epsilon_0 A}$

$E=\frac{(5.31-3.54) \times 10^{-7}}{(8.85 \times 10^{-12})(2.00 \times 10^{-1})}$

$E=1.00 \times 10^5 \ V/m$

Applications of the Effect of Dielectric on Capacitance

1. Capacitors in Electronic Circuits
   Dielectric materials are used in capacitors to increase capacitance without increasing the size of the plates. This allows compact electronic devices such as smartphones, laptops, and power supplies to function efficiently.

2. Energy Storage Systems
   By inserting a dielectric material, capacitors can store more electrical energy. This is useful in energy storage devices, flash cameras, and pulsed power systems.

3. Tuning Circuits (Radio and TV)
   Dielectrics help adjust capacitance in tuning circuits. By changing the dielectric constant, circuits can select different frequencies, which is essential in radios, televisions, and communication devices.

4. Capacitive Sensors
   Dielectric materials are used in sensors to detect changes in distance, pressure, humidity, or position. When the dielectric changes, the capacitance changes, which can be measured and used in applications like touch sensors and proximity sensors.

5. Capacitive Touchscreens
   In smartphones and tablets, the human finger acts as a dielectric. When you touch the screen, it changes the capacitance, allowing the device to detect the touch location.

6. Insulation in Electrical Systems
   Dielectric materials not only increase capacitance but also act as insulators, preventing current leakage and improving the safety and efficiency of electrical devices.

7. Variable Capacitors
   Some devices use movable dielectric materials to vary capacitance. These are used in tuning circuits and signal processing equipment.

8. High-Voltage Equipment
   In power transmission and high-voltage systems, dielectric materials help increase capacitance and prevent electrical breakdown, ensuring stable operation.

Conclusion – Effect of Dielectric on Capacitance

The effect of a dielectric on capacitance is fundamental in understanding how capacitors can be enhanced for practical use. When a dielectric material is inserted between the plates of a capacitor, the capacitance increases by a factor equal to the relative permittivity of the material. This happens because the dielectric reduces the effective electric field, allowing more charge to be stored for the same potential difference. The relationship is given by:

$C=\frac{\epsilon_r \epsilon_0 A}{d}$

This shows that capacitance depends not only on the geometry of the capacitor but also on the properties of the dielectric material. In real-world applications, the use of dielectric materials makes capacitors more efficient, compact, and capable of storing greater amounts of energy. This principle is widely applied in electronic circuits, energy storage devices, sensors, and communication systems.

Overall, understanding the effect of dielectric on capacitance provides a strong foundation for both theoretical physics and modern technological applications, highlighting the importance of material properties in electrical systems.

Frequently Asked Questions (FAQ) – Effect of Dielectric on Capacitance

1. What is a dielectric material?
A dielectric material is an insulating substance that does not conduct electricity but can be polarized in the presence of an electric field.

2. What happens to capacitance when a dielectric is inserted?
The capacitance increases. It becomes εᵣ times greater than the capacitance in vacuum.

3. What is the formula for capacitance with a dielectric?
The capacitance is given by:

C = εᵣ ε₀ A / d

where εᵣ is the relative permittivity, ε₀ is the permittivity of free space, A is the plate area, and d is the distance between the plates.

4. Why does capacitance increase with a dielectric?
Because the dielectric reduces the effective electric field between the plates, allowing more charge to be stored at the same potential difference.

5. What is relative permittivity (εᵣ)?
Relative permittivity is a measure of how much a dielectric material increases the capacitance compared to vacuum.

6. Does the potential difference change when a dielectric is inserted?
It depends on the condition:

• If the capacitor is connected to a battery, the potential difference remains constant.
• If it is isolated, the potential difference decreases.

7. What happens to the electric field when a dielectric is inserted?
The electric field decreases because the dielectric opposes the original field through polarization.

8. What are some examples of dielectric materials?
Common examples include air, glass, plastic, paper, and mica.

9. Where is this concept used in real life?
It is used in capacitors, electronic circuits, sensors, touchscreens, and energy storage systems.

10. Can all materials act as dielectrics?
No, only insulating materials can act as dielectrics. Conductors cannot be used as dielectric materials.

Share :