Electric Charges• Matter is made up of tiny particles

called atoms. In the centre of each atom, there is a nucleus. This is made up of protons and neutrons. Orbiting round the nucleus are electrons.

• Protons are positively charged. Electrons are negatively charged. The size of charge in an electron is equal to that in a proton. Neutrons have no charge.

Unit of charge• Charge is measured in

coulomb C.

• Charge of a proton

= 1.6 x 10-19 C

• Charge of an electron

= -1.6 x 10-19 C

Electric force• Like charge repel;

+ + F Q1 Q2 F

– –

F Q1 Q2 F

+ –

Q1 F F Q2

• Unlike charge attract.

Coulomb’s experiment• A quantity of charge, Q1, on a

fixed sphere repels a like charge, Q2, on a mobile sphere. This mobile sphere is attached to the end of a rod that is free to rotate because the rod is suspended by a fiber.

• The repulsion between the two charges pushes Q2 away a distance, r, until increasing tension in the twisting fiber balances the repulsion.

• The repulsion between the two charges equals the amount of force the twisted fiber exerts, which is known for any given angle.

Q1

Q2

fibre

suspension head

Coulomb’s law• Coulomb’s law states that the force between two point charges is

directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

r

Q2Q1

F21F12

FFF 1221

221

r

QQF

221

r

QQkF or

(Action and reaction)

Permittivity• The force between two charges also depends on the medium between

them.• If the charges are situated in an insulating medium instead of a vacuum,

the force between them will be reduced.• A medium is said to have permittivity; a material with high permittivity

is one which reduced the force between two charges greatly compared with the vacuum value.

• The force between two charges is expressed as

where is the absolute permittivity of the medium in which the charged are placed.

• In free space (vacuum), = 0 = 8.85 x 10-12 C2 N-1 m-2 where 0 is called the permittivity of free space.

221

4

1

r

QQF

Relative permittivity• Relative permittivity r is introduced to compare the

permittivity of different media easily. • The permittivity of a medium is the product of the permittivity

of free space and its relative permittivity. i.e. = 0 r

Medium Relative permittivity rVacuum (free space) 1

Air 1.0005

Paraffin 2.1

Mica 6

Acetone 27

Methyl alcohol 28

Water 81

• Note: Since the permittivity of air is very close to that of free space (relative permittivity of air is 1.0005 ≈ 1), the force between the charges in air can be

approximated as 2

21

04

1

r

QQF

TAS Lab

CycleGroup

20(10/4)

21(22/4)

22(30/4)

1 B7 C8 A2

2 B7 C8 A2

3 A2 B7 C8

4 A2 B7 C8

Example 1 Coulomb’s lawFind the magnitude of the electric forces in the following cases.(a) Two +1C point charges separated by 1 mm in air.(b) Two point charges, + 0.1 C and – 0.2 C, are separated by 20

cm in water.

• Solution:

(a) F = [1/(40)](1 x 1)/(0.0012) = 8.99 x 1015 N

(b) F = [1/(40r)](0.1 x 0.2)/(0.22) = 5.55 x 107 N

Example 2 Coulomb’s law and centripetal forceIn a hydrogen atom, an electron moves around a nucleus in a circular path. Estimate the speed of the electron given the following information.Mass of an electron = 9.31 x 10-31 kgCharge of an electron = 1.6 x 10-19 CRadius of the orbit = 5 x 10-11 mSolution:

• Centripetal force = electric force

= [1/(4o)] (1.6 x 10-19 )(1.6 x 10-19 )/ (5 x 10-11 )2

= 9.2076 x 10-8 N• By F = mv2/r

9.2076 x 10-8 = (9.31 x 10-31)v2 / (5 x 10-11) v = 2.22 x 106 ms-1

speed of an orbiting electron is 2.22 x 106 ms-1

Example 3 Addition of electric forces Three charged particles A, B and C are fixed at the corners of a square of length a as shown below. What is the resultant force acting on particle B? [Let k = 1/ (40)]

• F = k(1 x 1)/(a2) = k/a2

• Magnitude of resultant force = √5 F = √5 k/a2

• Direction of resultant force: = tan -1 (2/1) = 63.4o

The resultant force makes an angle 63.4o with AB.

A (+ 1 C)

C (+ 2 C)

B (+ 1 C)

2F

1F

√5 F

a

a

Electric field• An electric field is a region where an electric charge

experiences a force. Electric charges are sources in setting up electric field which in turn affect other charges placed inside the field.

• Electric field is represented by field lines and the field lines cannot cross

• The direction of the field lines shows the direction of the electric force acting on a positive test charge.

• Where the lines are parallel and uniform spaced, the field is uniform.

Electric field patterns

An isolated positive charge An isolated negative charge

+ –

The electric field patterns for different arrangement of charges are shown below.

Electric field patterns

A pair of opposite A pair of positive A pair of parallel charges charges plate with opposite

charges

The electric field patterns for different arrangement of charges are shown below.

Electric field strength• The density of field lines is proportional to the electric field

strength.• Where the field lines are closely spaced, the field is strong.

Where they are widely spaced, the field is weak.

weak field

strong field

Electric field strength• If a (positive) test charge Q0 placed at a point experience a

electric force F, then the electric field strength E at that point is defined to be the force experienced per unit charge.

Charge

ForcestrengthfieldElectric

0Q

FE or

Mass

ForcestrengthfieldnalGravitatio

m

Wg or

+Q0

F

Note:• 1. The direction of E is same as that of the force F.• 2. Q0 should be extremely small so that it would not

disturb the original electric field.

+Q0

F

Electric field strength of a positive point charge

• By Coulomb’s law, the electric force acting on the test charge +Q0 is

• The electric field at point P is

+Q

+ +r

+Q0

20

4

1

r

QQF

20

4

1

r

QQF

0Q

FE 24

1

r

QE

i.e.

Electric field strength of a negative point charge

• By Coulomb’s law, the electric force acting on the test

charge – Q0 is

• The electric field at point P is

–Q

+ –

r

+Q0

20

4

1

r

QQF

20

4

1

r

QQF

20

4

1

r

QQF

0Q

FE

24

1

r

QE

i.e.

Note:Note: EE depends only on depends only on QQ which produces the field, and not on the which produces the field, and not on the value of the test charge value of the test charge QQ00..

Electric field strength• If a (positive) test charge Q0 placed at a point experience a

electric force F, then the electric field strength E at that point is defined to be the force experienced per unit charge.

Charge

ForcestrengthfieldElectric

0Q

FE or

Mass

ForcestrengthfieldnalGravitatio

m

Wg or

+Q0

F

Electric field strength of a positive point charge

• By Coulomb’s law, the electric force acting on the test charge +Q0 is

• The electric field at point P is

+Q

+ +r

+Q0

20

4

1

r

QQF

20

4

1

r

QQF

0Q

FE 24

1

r

QE

i.e.

Example 4A negative point charge of -6p C is placed at point A in vacuum. Calculate the electric field strength at point P, where the distance between A and P is 2 m. [Take 1/(40) = 9 x 109 Nm2C-2]

• Solution:

Electric field strength = [1/(40)](Q/r2)

= 9 x 109(6 x 10-12)/22

= 0.0135 NC-1

Example 5ABC is an equilateral triangle of side a. A positive and a negative point charge are placed at B and C respectively. The magnitude of each of the point charge is Q. Find the resultant electric field strength at A. [Let k = 1/(40)]

A

B+Q

C–Q

E

Electric field strength E due to +Q= [1/(40)](Q/r2)= kQ/a2

Electric field strength E due to –Q = kQ/a2

Resultant E field strength = kQ/a2

a

aa

E

Er

Note: E is a vector. The Note: E is a vector. The resultant field must be obtained resultant field must be obtained by using vector addition.by using vector addition.

Electric field strength of a charged conducting sphere

• The charges in a charged conductor repel one another. Therefore, they distributed uniformly over the surface of the sphere.

• Electric field is always perpendicular to the surface of the conducting sphere.

• A charged sphere behaves like a point charge at its centre.

+

+

+

+

+

+

+ +

A charged conducting sphere

+

A point charge

204

1

r

QE

Inside the sphere, the electric field is zero.

Outside the sphere, (r a ), the electric field is

At the surface of the sphere, (r = a),

is known as the charge per unit area of the surface of the conductor or the charge density.

or where 24 a

Q

2

04

1

a

QE

0

E

204

1

r

QE

ra

E

o

Comparison between gravitational field and electric field

Gravitational field Electric field

Symbol g E

Objects affected Objects with mass Objects with charge

Meaning

Unit N kg-1 N C-1

Vector Vector

m

Wg

Q

FE

Electric potential energy and gravitational potential energy

• The gravitational potential energy of a mass is increased if it is moved to a higher level (against the gravitational field).

• Similarly, the electric potential energy of a positive charge is increased if it is moved from A to B (against the electric field). When the charge is allowed to return from B to A, the electrical potential energy gained previously is changed into kinetic energy.

• In lower forms (F4 – F5), the gravitational potential energy at ground level is taken as zero.

• Now, theoretically, the electric potential energy of a charge at infinity is taken as zero. (Remember: the p.e. of gas molecules)

The mass gains gravitational potential energy

groundground

++QQ

+++r

++QQ00

B Ax

Electric potential energy and electric potential

• The electric potential energy W of a charge Qo at a point in an electric field is defined as the work done (energy) to move the charge Qo from infinity to that point.

• The electric potential V at a point in an electric field is defined as the work done in moving a unit positive charge from infinity to that point.

Q

WV

Q

WV

Potential due to a uniform field E

• The potential difference across two points A and B is defined as the work done required to move a unit positive chargea unit positive charge from point A to point B.

• Consider two points A and B of distance d apart in the following uniform field.

The electric potential difference between A and B is given by V = Ed.

B Ax x

d

++

Q0

Proof:

Work done = F x s, F = EQ0

Put Q0 = 1 (unit charge)

Potential difference V

= Work done

= [E(1)]d = Ed

Example 6Electric potential is more commonly known as voltage. The potential across the two parallel plates is 5 V and the distance between the plates is 3 cm. (a) Find the magnitude of electric field strength set up by the charged plates.(b) If a proton is released from the positive plate, find its speed when it reaches the negative plate. Assume that the medium between the plates is vacuum.

Given that the mass of a proton is 1.67 x 10–27 kg and the charge of a proton is 1.6 x 10–19 C.

Solution:(a) By V = Ed

(5) = E(0.03)E = 167 Vm-1

Note: There are two units for E.(1) E = F/Q (unit : N C-1)(2) E = V/d (unit: V m-1)

Example 6Electric potential is more commonly known as voltage. The potential across the two parallel plates is 5 V and the distance between the plates is 3 cm. (a) Find the magnitude of electric field strength set up by the charged plates.(b) If a proton is released from the positive plate, find its speed when it reaches the negative plate. Assume that the medium between the plates is vacuum.

Given that the mass of a proton is 1.67 x 10–27 kg and the charge of a proton is 1.6 x 10–19 C.

Solution:(b) Electric potential energy stored is changed into

kinetic energyQV = ½ mv2

5(1.6 x 10-19) = ½ (1.67 x 10-27)v2

v = 3.10 x 104 ms-1

Potential due to a point charge

+Q

++ +

r

+Q0

x

+++Q0

20

4

1

x

QQF

Displacementr

Force

The potential of a point at a distance of r from a positive point charge Q is

r

QV

4

1

Potential due to a point charge

20

4

1

x

QQF

Displacementr

Force

Electric potential V = Work done in moving a unit positive charge from infinity to a point= area under the force – displacement graph

(– sign indicates the displacement is in the opposite direction as the field)

=

r

dxF=

r

dxx

Q24

1

r

QV

4

1r

Q

4

1=

Example 7Find the potential at points A and B due to two point charges X and Y, 1 m apart in air and carrying charges of 2 x 10-8 C and -2 x 10-8 C respectively. • For point A:

Potential due to X

= [1/(40)](QX/1)= 9 x 109 (2 x 10-8) / 1 = 180 VPotential due to Y

= [1/(40)](QY/2)= 9 x 109 (-2 x 10-8) / 2 = -90 V

• Potential at point A = 180 + (-90) = 90 V

Note: Potential is a scalar.Thus, the resultant can be obtained by simple addition.

A

B

X Y

1 m

1 m 1 m

1 m

Example 7Find the potential at points A and B due to two point charges X and Y, 1 m apart in air and carrying charges of 2 x 10-8 C and -2 x 10-8 C respectively.

• For point B:

Potential due to X

= [1/(40)](QX/1)

= 9 x 109 (2 x 10-8) / 1 = 180 V

Potential due to Y

= [1/(40)](QY/1)

= 9 x 109 (-2 x 10-8) = -180 V

• Potential at point B

= 180 + (-180) = 0 V

A

B

X Y

1 m

1 m 1 m

1 m

Electrical potential energy stored in a system of charges

+Q1

+

r

+Q2

+

The electrical potential energy stored when two positive charges Q1

and Q2 at a separation of r is

Proof:

r

QQW 21

4

1

The electrical potential energy stored in the system = work done to move the charges from infinity to the positions required = work done to move Q2 from infinity while Q1 is fixed

rr

dxx

QQFdx

221

4

1

r

QQ 21

4

1

Example 8Two positively-charged balls are tied together by a string. One ball has a mass of 30 g and a charge of 1 C; the other has a mass of 40 g and a charge of 2C. The distance between them is 5 cm. Initially they are at rest, but when the string is cut they move apart. What are their speeds v1 and v2 when they are very far apart?

r

QQ 21

04

1

Solution:

Initial potential energy of the system= k Q1Q2/r

= 9 x 109 (1 x 10-6)(2 x 106) / 0.05

= 0.36 J When the balls are very far apart, the potential energy is zero and final kinetic energy = ½ mv1

2 + ½ mv22

Energy is conserved, so the final kinetic energy = initial potential energy ½ (0.03)v1

2 + ½ (0.04)v22 = 0.36

⇒3v12 + 4v2

2 = 72 --- (1)

By conservation of momentum,m1v1 = m2v2

0.03v1 = 0.04 v2

⇒ v1 = 4v2/3 --- (2)

Solving (1) and (2): v1 = 3.70 ms-1 and v2 = 2.78 ms-1.

30 g 40 g 30 g 40 g

Potential due to a charged sphere

• Inside the sphere, electric field is zero. Thus, the electric potential is constant, but not zero.

• The field at any point outside the

sphere is exactly the same as if the

whole charge were concentrated at the

centre of the sphere. i.e.

• The potential at the surface is

where a is the radius of

the sphere.

r

QV

4

1

a

QV

4

1

Equipotential lines

• Equipotential lines are connected lines of the same potential. • The equipotentials are always perpendicular to the field lines.• The density of the equipotentials represents the strength of the

electric field.• The equipotentials never cross each other.• If a charge moves along an equipotential line, no work is done; if

a charge moves between quipotential lines, work is done.

equipotential lines

electric field lines

Field and equipotential lines Field and equipotential lines for a positive point chargefor a positive point charge

Field and equipotential Field and equipotential lines for a uniform fieldlines for a uniform field

400 V300 V200 V100 V0 V (arbitrary zero)

200 V

150 V

100 V

Equipotential surfaces and field lines

• The equipotentials are always perpendicular to the field lines.

• The density of the equipotentials represents the strength of the electric field.

• The equipotentials never cross each other.

A conducting Material in an Electric Field

• Consider a pair of oppositely charged plates which established a uniform field between them.

+ -

conductorV/V

0 x/m

E/V m-1

0 x/m

Electrostatic Shielding

• The field inside the hollow metal box is zero. • A conducting box used in this way is an effective device for

shielding delicate instruments and electronic circuit from unwanted external electric field.

• The inside of a car or an airplane is relatively safe from lightning.

+

-

Influence on a charged conductor

Consider an isolated positively charged conductor.The potential is decreased by an earthed object, an uncharged object or a negatively charged object.The potential is increased by a positively charged object.

an earthed object

a neutral object

a negatively charged object

a positively charged object

Flame probe investigation of electric potential

• A flame probe is used to measure the potential in space.

• The ions in the fame neutralize the charge on the needle. Therefore, there is no excessive charge on the needle. i.e. the probe will not influence the potential to be measured.

• the potential can be read from the deflection of the leaf