Topic 3 Electric Potential Energy and Capacitance

TTooppiicc 33 Electric Potential Energy and Capacitance

LEARNING OUTCOMES By the end of this topic, you should be able to:1. Define the electric potential and the electric potential difference between

two points and apply them to related problems; 2. Distinguish between the electric potential and the potential difference; 3. Calculate the electric potential energy associated with a group of

charges; 4. Describe the equipotential surface as a surface on which the potential

has the same value at every point; 5. Describe the principle of the capacitor as an energy-storing device and

define the capacitance; 6. Determine the equivalent capacitance of capacitors arranged in series,

parallel or the combination of both; 7. Determine the electric energy stored in a charged capacitor in the

absence of the dielectric material; and 8. Describe the effects of a dielectric placed between the plates of a

parallel-plate capacitor.

TOPIC 3 ELECTRIC POTENTIAL ENERGY AND CAPACITANCE

34

INTRODUCTION

The concept of electrical potential energy is the same as the concept of gravitational potential energy or elastic force of a spring that you have studied in mechanics. It is of great value in the study of electricity. It enables us to define a scalar quantity known as electric potential. When a charged particle moves in an electric field, the field exerts a force that can do work on the particle. This work can be expressed in terms of electrical potential energy. The electric potential between two points A and B in a uniform electric field is known as the potential difference In electrical circuits, the potential difference is referred to as the voltage and can be measured using an instrument known as a voltmeter. The concepts of potential difference and voltage are of great importance in our daily life, since all devices powered by electricity require certain values of voltage in order to do work. A capacitor is a device that stores electric potential energy and electric charge. The device consists of conductors separated by an insulator or a dielectric material. Capacitors have many applications such as filters of power supply, energy-storing devices in flash units for photography, tuning frequency devices for radio and T.V, and devices to eliminate sparking in car ignition systems. In this chapter we will study the fundamental properties of capacitors including the definition of capacitance, the equivalent capacitance of a parallel or series combination of capacitors, energy stored in a charged capacitor, the effect of dielectrics to the properties of capacitors.

TOPIC 3 ELECTRIC POTENTIAL ENERGY AND CAPACITANCE 35

ELECTRIC POTENTIAL ENERGY

The electrostatic force that we studied in Topic 2 is a conservative force. By that, we mean that the work done by this force in moving a charge depends only on the initial and final positions only and not on the path connecting these positions. Since the force is conservative, we are able to define the electric potential energy associated with this force. Let’s now consider a positive charge q that is placed in a constant electric field E between two oppositely charged plates, Figure 3.1.

Figure 3.1: The work done by a charge as it moves in an electric field

A B

E

Gravitational potential energy is the energy stored in an object due to its vertical position above the Earth. The amount of gravitational potential energy stored in an object depends on (i) the mass the object possesses, and (ii) the height to which it is raised. How would you describe electric potential energy in analogous comparison with gravitational potential energy? What are the electrical equivalents of the gravitational mass and height?

SELF-CHECK 3.1

3.1


36

Let the magnitude of the electric field be E. Now suppose we move this charge slowly from point A to point B along a straight line in the direction of E. Let the distance between A and B be equal to x. As the charge moves from A to B, work is done by the electric field on it. Since the electric force acting on the charge is F = qE, the amount of work that is done on the charge as it moves from A to B is given by:

ABW Fx qEx= = (3.1) The electrostatic force causes the charge to accelerate to the right, gaining kinetic energy. As the charge gains kinetic energy, it loses an equal amount of electrical potential energy. We define the work done by a conservative force to be equal to the negative of the change in the potential energy. Thus, the change in the electrical potential energy

UΔ is:

ABU W qExΔ = − = − (3.2) In general, the electric potential energy U of a test charge q at distance r from the charge Q may be expressed as:

qQU kr

= (3.3)

The potential energy of a positive charge is higher at a point nearer to the charge Q that produced the electric field E. When the potential energy increases, the kinetic energy of the particle decreases. On the other hand the potential energy of negative charge is higher at a point away from the charge Q, and, accordingly, the kinetic energy decreases away from the charge source Q. According to the work-energy theorem, the change in kinetic energy during any displacement is equal to the work done on the particle by the potential energy. Thus, the total mechanical energy (kinetic plus potential) is conserved and can be expressed as:

Ka + Ua = Kb + Ub (3.4) This is equivalent to the case of a particle moving under the influence of the gravitational field, which you studied in Topic 4 of the SBPH2103 course. There we saw that the gravitational potential energy of the particle increases with height while the kinetic energy decreases.


ELECTRIC POTENTIAL AND POTENTIAL DIFFERENCE

The electric potential is commonly known as the potential. The electric potential V is defined as the potential energy U per unit charge q:

UV =q

(3.5)

Since the potential energy and charge are both scalars, it necessarily follows that the electric potential is also a scalar quantity. The potential difference (p.d) between 2 points, A and B, B AV V− , is defined as the change in potential energy of a charge q as it is moved between these two points divided by the magnitude of the charge.

B AUV V VqΔ

Δ = − = (3.6)

Notice that the potential difference depends on the electric potential of the initial and final points. From equation (3.6), the unit of V is seen to be J/C. But because the electric potential is an important quantity in physics, we give this combination of units a very special name, i.e., the Volt, where 1 Volt = 1V = 1 J/C. A unit of energy associated with the volt is the electron-volt (eV), a unit commonly used in atomic and nuclear physics. The eV is defined as the potential energy that an electron (having charge 191.6 10e −= × C) will gain when it moves through a potential difference of 1 V. When this happens, the electron gains an amount of electric potential energy that has a magnitude of:

( )( )19 191.6 10 C 1V 1.6 10 JU q V − −Δ = Δ = × = × (3.7)

3.2

What is the difference between: (a) electric potential and electric field, (b) electric potential and electric potential energy?

SELF-CHECK 3.2


38

THE ELECTRIC POTENTIAL OF A POINT CHARGE

The electric potential V at a point P a distance r from a point charge Q is defined as:

QV kr

= (3.8)

If there are more charges, then the electric field at point P can be calculated by adding up the electric potentials due to each charge:

31 2

1 2 3

i

i

Q QQ QV k .... kr r r r

⎛ ⎞= + + =⎜ ⎟

⎝ ⎠∑ (3.9)

Example 3.1

An electric force moves a positive charge of magnitude 100μC from point A to point B and does 5 mJ of work on the charge.

(a) Find the difference between the electric potential energies of the charge at A and B.

(b) Find the potential difference between points A and B.

(c) Which point is at a higher potential? Solution (a) The change in the electric potential energy is given by Equation 3.2: 5.00mJB A ABU =U -U = - WΔ = −

(b) The potential difference between the points is

3

6

5 10 50V100 10A B

UV Vq

−

−

Δ − ×− = = = −

×

The electric potential at a point is zero. Is it possible for the electric field at that point to be nonzero? If yes, give an example, if no explain why not.

SELF-CHECK 3.3

3.3


(c) Point B has the higher potential Example 3.2 What is the electric potential at point P, located at one of the corner of a rectangle with side dimensions of 3 mm x 4 mm as shown in Figure 3.2?

Figure 3.2

Solution

Apply Eq 3.9 i 31 2

i 1 2 3

Q QQ QV = k = k + +r r r r

⎛ ⎞⎜ ⎟⎝ ⎠

∑

-6 -6 -6

9-3 -3 -3

-6.0×10 4.0×10 10.0×10= 9.0×10 + +3.0×10 4.0×10 5.0×10

V⎛ ⎞⎜ ⎟⎝ ⎠

∑

( )6 6= 9.0×10 -2 +1+ 2 = 9.0×10V Volt

Example 3.3: Electric potential due to point charges A particle of charge 1 6μCq = + is located on the x-axis at the point 1 5.1cm.=x A second particle of charge 2 5μCq = − is placed on the -axis at 2 3.4cmx = . What is the electric potential at the origin, i.e., at 0cmx = ? Solution The electric potential at the origin due to the first charge 1 6μCq = + is

9 661

1 21

9 10 6 10 1.06 105.1 10

kqVx

−

−

× × ×= = = ×

× V

P

+ 10 μC –6 μC

+4 μC

3 mm

4 mm


40

Likewise, the electric potential at the origin due to the second charge 2 5μCq = − is

( )9 662

2 22

9 10 5 101.32 10

3.4 10kqVx

−

−

× × − ×= = = − ×

× V

The net potential at the origin is simply the algebraic sum of these two potentials. Thus,

51 2 2.6 10 VV V V= + = − ×

RELATION BETWEEN E AND V

Let’s again consider the situation of a positive charge q placed in a constant electric field E between two oppositely charged plates. Let the magnitude of the electric field be E. We already know from equation 3.2 how to calculate the change in the electric potential energy of the charge as it is moves between 2 points, A and B, that lie along a straight line. But what is the change in the electric potential of this charge? We can get the answer immediately by using equation (3.7):

U qExV Exq qΔ −

Δ = = = − (3.10)

This means that we can take the electric potential between the plates as

platesV Ex= − (3.11)

3.4

1. A 20C charge is brought from infinity (where the electric potential is taken to be zero) to point A in an electric field. The work done in this process is 20J. What is the electric potential at A?

2. Four point charges of 2.0 nC, -2.4 nC, 3.2 nC and 2.5 nC are placed at

the vertices of a square 1cm on a side. What is the electric potential located at the centre of this square?

EXERCISE 3.1


Note that x points from the +ve plate to the –ve plate. If we are given the potential difference between the plates and the distance between the plates ( x = d), the magnitude of the electric field E can always be found from

VEdΔ

= (3.12)

Example 3.4 The spark plug in your car consists of two metal plates that are separated by a small distance. Take d = 0.1m. When an electric spark jumps between the plates, the magnitude of the electric field is close to 650 10 V/m× . Find the potential difference between the plates. Solution

The magnitude E of the electric field is given by equation (3.12) as VEdΔ

= ,

where ΔV is the potential difference between the two metal plates and d is the distance between them. Thus the potential difference is ( )( )6 650 10 0.1 5 10 VV EdΔ = = × = ×

Example 3.5 Two parallel plates of separation 0.25 cm in vacuum are connected to 100-V battery terminals. An electron is released from the negative terminal. Find

(a) The electric field between the plates

(b) The work done by the electron in eV

(c) The velocity of the electron

Solution

(a) The electric field -3

100= = 40,000 V/m2.5×10

VE =d

(b) Work done 19 171 6 10 100 1 6 10− −= × = ×W = qV ( . )( ) . J or 100 eV (c) The kinetic energy is equal to the potential energy in eV; 21

e e2 m v = eV


42

Hence the velocity of the electron is given by

ee

2eVv =m

-176 -1

-31 -31

2×(100eV) 2×(1.6×10 J) = 5.93×10 ms9.11×10 kg 9.11×10 kg

==

EQUIPOTENTIAL SURFACES

An equipotential surface is a surface on which the electric potential V is the same at every point. Figure 3.3 shows the cross-sections of equipotential surfaces of parallel conducting plates, a single isolated point charge, and an electric dipole. Note that for each of these cases, the field lines and equipotential surfaces are always mutually perpendicular to each other. The equipotential surfaces are also closer together in the region where the electric field is stronger.

On equipotential surfaces, no point can be at two different potentials. So equipotential surfaces for different potentials can never touch or interact. If a charge moves on the equipotential surface, the electric field can do no work there, since the potential difference between any two points on this surface is zero.

3.5

1. The electric field between two metal plates 10 cm apart is 500 V/m. (a) What is the potential difference that must be applied across the

conductors? (b) What is the force on a charge of 10 nC released in this field? (c) What would be the kinetic energy if the charge has moved 5 cm

in the field? 2. A 8.0-kV DC power supply is connected between two conducting

plates10 cm apart. An electron is released from one plate and undergoes a displacement to the opposite plate. Find (a) The magnitude of the electric field (b) The work done and (c) The speed of the electron.

EXERCISE 3.2


Dashed lines are equipotential lines while solid lines are electric field lines.

Figure 3.3: Equipotential surfaces

CAPACITANCE

A capacitor is designed to store electric charges. Any two conductors, such as a pair of parallel metal plates, separated by an insulator (a vacuum or a dielectric) form a capacitor. In circuit diagrams a capacitor is represented by either of these symbols shown in Figure 3.4.

Figure 3.4: Symbols of a capacitor in an electrical circuit To charge a capacitor, we connect it with wires to opposite terminals of a battery. Once the charges +Q and -Q are established on the plates of the capacitor, the battery is disconnected. Suppose the potential difference between the parallel plate conductors is V. The capacitance of the capacitor is defined as the ratio of the magnitude of charge on one side of the plate to the magnitude of the potential difference between the plates.

QC =V

(3.13)

3.6


44

The SI unit of capacitance is the farad (F), which is coulomb per volt.

1 F = 1 C/ V. The farad is a very large unit of capacitance. In practice, typical devices have capacitances ranging from microfarad ( Fμ ) to picofarad (pF), depending on the application.

CALCULATION OF CAPACITANCE

3.7.1 Parallel-plate capacitors

Two parallel plates of equal area A are separated by a distance d and oppositely charged with charge Q as shown in Figure 3.5. The charge on either plate is Q. Let the p.d between the plates be V

Figure 3.5: A parallel-plate capacitor By definition, the capacitance is

0ε AQC = =V d

(3.14)

where 0ε is the permittivity of free surface. As you can see from Equation (3.14), the capacitance of a parallel-plate capacitor is proportional to the area and inversely proportional to the separation of the plates.

3.7


Example 3.6 A parallel-plate capacitor has an area of 24100.5 m−× and a plate separation of 1.0 mm. Find its capacitance. Solution Solving equation 3.14 for the capacitance, we get

-12 -4 2

-12-3

(8.85×10 F/m)(5.0×10 m ) = 4.43×10 F = 4.43pF1.0×10 m

0ε AC = =d

CAPACITORS IN SERIES AND PARALLEL

Several capacitors can be connected in a circuit by many combinations. The simplest combinations are a series connection and a parallel connection. The equivalent capacitance is then calculated according to the series and parallel rules. In situations containing both series and parallel connections, we need to reduce the circuit in steps to obtain the equivalent capacitance.

3.8

1. A parallel plate capacitor has plates of cross-sectional area 20.4 m . The distance between the plates is 2 mm. If a 25 kV p.d is applied to the capacitor, calculate

(i) The capacitance

(ii) The charge on each plate. 2. The plates of a parallel-plate capacitor in vacuum are 2.5 mm apart and

2.0 cm2 in area. A potential difference of 1.0 kV is applied across the capacitor. Calculate the

(a) The capacitance,

(b) The charge on each plate, and

(c) The magnitude of electric field between the plates.

EXERCISE 3.3


46

3.8.1 Series Combination

Figure 3.6 shows a battery with the potential V is connected to three capacitors in series by conducting wires. In a series connection the magnitude of charge in all individual capacitors is the same and equal to Q. The potential different across each capacitor is

Figure 3.6: (a) A series combination of three capacitors and (b) the equivalent capacitance

of the circuit The total potential difference V on the equivalent capacitor is

1 2 31 2 3

Q Q QV = V +V +V = + +

C C C, or

1 1 1

1 2 3

V= + +

Q C C C

The equivalent capacitance eqC = Q/V or

1 1 1 1

eq 1 2 3= + +

C C C C (3.15)

Therefore, the equivalent capacitance of a series connection equals the sum of the reciprocals of the individual capacitances.


3.8.2 Parallel Combination

Figure 3.7 shows three capacitors that are connected in parallel by conducting wires to the terminals of a battery with the potential V. In a parallel connection the potential difference for all individual capacitors is the same and equal to V. The charge on each capacitor is

1 1 2 2 3 3Q = C V Q = C V Q = C V The total charge Q on the equivalent capacitor is

1 2 3 1 2 3Q = Q + Q +Q = (C + C +C )V

1 2 3

Q= C +C +C

V

The equivalent capacitance eqC = Q/V or

eq 1 2 3C = C +C +C (3.16) Therefore, the equivalent capacitance of a parallel connection equals the sum of the individual capacitances.

Figure 3.7: (a) A parallel combination of three capacitors and (b) the equivalent capacitance of the circuit.

Example 3.7 3 capacitors of capacitances 1 22 2C F ,C F= μ = μ and 3 1C F= μ are connected to a 100V supply as shown in Figure 3.8. Find

(a) the equivalent capacitance of the combination

(b) the charges and voltages on each capacitor.


48

Figure 3.8 Solution (a) In Figure 3.8(a), the capacitance of the parallel combination of 2C and 3C is

C′ = 2C + 3C =3 F.μ

Now C′ is connected in series with 1C . See Figure 3.8(b). Hence the equivalent capacitance C of the entire combination is given by

1

1

1

1 1 1

2 3 6 1 22 3 5

C C C

C CC . FC C

= +′

′ ×= = = = μ

′+ +

(b) Let 1 2Q ,Q and 3Q be the charges on 1 2C ,C and 3C respectively. The total

charge Q stored in this combination is

1 2 3 1 2 100 120Q Q Q Q CV . F V C= = + = = μ × = μ

Therefore, the voltage across 1C is 11

1

60QV VC

= =


Since 2C is parallel to 3C , the voltage across 2C = voltage across 3C , i.e.

2 3 100 60 40V V V V V .= = − = Hence 2 2 2 80Q C V C.= = μ 3 3 3 40Q C V C.= = μ

Example 3.8 The figure below shows a circuit consisting of 4 capacitors, each having a capacitance of 10μF . What is the effective capacitance?

Solution The capacitance of the parallel combination of 1C and 2C is

C′ = 1C + 2C =10 10 20F F F.μ + μ = μ Now C′ is connected in series with 3C and 4C :

3C

C′

4C

C1 C2

C3

C4


50

Hence the equivalent capacitance C of the entire combination is given by

3 4

1 1 1 1

1 1 1 110 10 204

C C C C

CC F

= + +′

= + +

= μ

ENERGY STORAGE IN CAPACITORS

Many applications of capacitors depend on their ability to store energy in the form of the electric potential energy. Electric energy can be stored in capacitors in the form of charges. The electric potential energy stored in a capacitor is equal to the amount of work required to charge the capacitor. Suppose a capacitor with capacitance C is charged to the final charge Q and the final potential difference across the capacitor is V. It can be shown that the total energy stored U by the capacitor is given by:

3.9

(a) Find the equivalent capacitance of the network shown below.

(b) Determine the charge and voltage across each capacitor. 3 3C F= μ , 4 4C F= μ , 6 6C F= μ

EXERCISE 3.4


2

2QUC

= (3.17)

If we use the relationQ CV= ,then the total energy can also be expressed as:

212

U CV= (3.18)

Example 3.9 What voltage is required to store 20 J of energy in a 370μF capacitor ? Solution

U = 20 J , C = 370μF . From Eq. 3.18,

6

2 2 20 329V370 10

UVC −

×= = =

×

Try this analogy:

Compare the flow of electricity with the flow of water out of a tank. If the tank is analogous to a capacitor (tank stores water, capacitor stores charges), use this water tank analogy to describe capacitor charging/discharging, electric potential, resistance etc. The more analogous pairings you can come up with, the better.

A 1.2μF capacitor charged to 3.0 kV. Compute the energy stored in the capacitor.

EXERCISE 3.5

ACTIVITY 3.1


52

DIELECTRICS

Dielectric materials are insulating materials. Most capacitors have dielectric made of plastic between their conducting plates. The dielectric serves three purposes, i.e. (a) To increase the capacitance of the capacitor,

(b) To increase the maximum possible potential difference between the plates after charging, and

(c) To provide electrical insulation in small separation between the conducting plates.

Suppose a parallel plate capacitor in vacuum, is fully charged to Q0 and potential difference between the plates isV0. The capacitance C0 is given by:

0 00

0

Q ε AC = =V d

(3.19)

With the capacitor still fully charged, a dielectric with dielectric constant 0ε κε= is inserted between the plates. The voltmeter reading decreases by a factor of κ , or

0VV =κ

(3.20)

where 1>κ is known as the relative dielectric constant. The capacitance C increases by a factor κ , or

0 00

0 0

Q QQC = = = κ = κCV V /κ V

0C = κC (3.21)

We can express the capacitance when the capacitor is filled with a dielectric as

00

κε A εAC = κC = =d d

(3.22)

3.10


We can also express the electric field and the energy stored in the electric field of the capacitor is filled with a dielectric as:

0EE =κ

(3.23)

0UU =κ

(3.24)

Example 3.10 Suppose the parallel-plate capacitor in Example 3.6 is connected to a power supply and charged to a potential difference 2,000 V. It then disconnected from the power supply and a sheet of dielectric material is inserted between the capacitor plates, completely filling the space between them. The voltmeter reading decreases to 500 V. Find

(a) The magnitude of charge on each plate,

(b) The capacitance after the dielectric is inserted,

(c) The dielectric constant,

(d) The permittivity of the dielectric Solution (a) 3 -9= (4.43pF)(2.0×10 V) = 8.86×10 C = 8.86nC0 0Q = C V .

(b) -9

-118.86×10 C= = 1.11×10 C = 0.111pC800 V

QC =V

(c) 2000V= = 2.5800V

0Vκ=V

(d) -12 -112.5×(8.85×10 F/m) = 2.213×10 F/m0ε= κε =


54

• The electric potential energy is the work done by the electric force on a

charged particle moving in an electric field.

• The potential energy for two point charges q and Q separated by a distance r is qQ

U = kr

.

• The potential energy is a scalar quantity.

1. A certain parallel-plate capacitor consists of two plates, each with area 2200cm , separated by a 0.40 cm air gap.

(a) Compute its capacitance

(b) If the capacitor is connected across a 500 V source, find the

(i) Charge on it

(ii) The energy stored, and

(iii) The value of the electric field between the plates.

(c) If a liquid with dielectric constant 1.59. is poured between the plates so as to fill the air gap, how much additional charge will flow onto the capacitor from the 500 V source?

2. The parallel plates of a capacitor have an area of 0.2 m2 and are 1 cm

apart. The capacitor has a potential difference of 3.0 kV after charging. A sheet of dielectric is inserted between the conductor plates and the potential different drops to 1.0 kV. Find

(a) The dielectric constant of the dielectric sheet

(b) The original capacitance and the final capacitance after insertion of dielectric

(c) The charge on each plate

(d) The initial and final energy stored in the capacitor

EXERCISE 3.6


• The electric potential is the potential energy per unit test charge q. The

potential due to many charges is Σ i

i

QV = k

r

• The potential difference is the work done by the electric force in taking a unit

test charge from point A to point B in the electric field due to a charge Q

• The electron-volt is the energy corresponding to a particle with a charge equal to that of an electron moving through a potential difference of 1 volt.

• An equipotential surface is a surface on which the electric potential has the same value at every point.

• A capacitor is any pair of conductors separated by an insulating material or dielectric.

• The capacitance C of a capacitor is defined as the ratio of the charge on either plate to the potential difference between the plates, or

• The parallel-plate capacitor is made with two parallel conducting plates, each with area A, separated at a distance d by a dielectric with dielectric constant

• When capacitors with capacitances 1 2 3, , ,C C C …are connected in parallel, the equivalent capacitance is equal 1 2 3eqC C C C= + + . When the capacitors are

connected in series, the equivalent capacitance is equal 1 2 3

1 1 1 1

eqC C C C= + + .

• The potential energy U required to charge a capacitor of capacitance C to a potential difference V and a charge Q is equal to the energy stored in the

capacitor and is given by 2 2 2

2 2Q CV QVU = = =

C.

• When the separation between the plates of the capacitor is filled with a

dielectric material of relative dielectric constant 1>κ , the new values of capacitance, charge, potential, and potential energy are affected by a factor of κ


56

Capacitance Capacitors Dielectrics

Electric potential Electric potential energy Equipotential surface

1. 5 capacitors of 10μF each are connected in (a) series, and (b) parallel. Find

the effective capacitance in each case. 2. 4 equal charges of magnitude 10 nC are located at the corners of a square of

edge length 5cm. Find the electric potential at the center of the square.

1. A proton is accelerated through a p.d of 1000V. What is the change in the

potential energy of the proton in electron-volts and in Joules? 2. A dielectric is inserted between the plates of a capacitor having constant

charge. What effect will this have on the following quantities?

(a) The capacitance

(b) The voltage between the plates

(c) The electric field between the plates

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Topic 3 Electric Potential Energy and Capacitance