Chapter 1: Electrostatics - YSL...

CHAPTER 1: ELECTROSTATICS

prepared by Yew Sze Ling@Fiona, KMPP

Chapter 1: Electrostatics

Subtopic C1 & C2 C3 & C4

1.1 Coulomb’s

Law States Coulomb’s Law,

2

0

2 4 r

Qq

r

kQqF

Sketch the electric force diagram

Apply Coulomb’s Law for a

system of point charges

1.2 Electric Field Define electric field strength,

0q

FE

Use electric field strength,

0q

FE

Use 2r

kQE for a point charge

Sketch the electric field strength

diagram

Determine the electric field

strength for a system of charges

1.3 Electric

Potential Define electric potential,

0q

WV

Define and explain equipotential

lines and surfaces of an isolated

charge in a uniform electric field

Use r

kQV for a point charge

and a system of charges

Calculate potential difference

between two points:

0q

WVVV initialfinal

Deduce the change in potential

energy between two points in

electric field: VqU

Calculate the potential energy for

a system of point charges:

23

32

13

31

12

21

r

qq

r

qq

r

qqkU

1.4 Charge in a

uniform

electric field

Explain qualitatively with the aid

of diagram the motion of charge

in uniform electric field

Use d

VE

for uniform E in:

stationary charge

charge moving

perpendicularly to the field

charge moving parallel to the

field

charge in dynamic

equilibrium



Introduction

The electrical nature of matter is inherent in atomic structure. An atom consists of a small,

relatively massive nucleus that contains particles called protons and neutrons. A proton has a mass

of 1.673×1027 kg, and a neutron has mass of 1.675×1027 kg. Surrounding the nucleus is a diffuse

cloud of orbiting particles called electrons. An electron has a mass of 9.11×1031 kg.

Like mass, electric charge is an intrinsic property of protons and electrons, and only two types of

charge have been discovered, positive and negative. A proton has a positive charge, and an

electron has a negative charge. A neutron has no net electric charge.

Experiment reveals that the magnitude of the charge on the proton exactly equals the

magnitude of the charge on the electron; the proton carries a charge +e, and the electron carries

a charge −e. The SI unit for measuring the magnitude of an electric charge is the coulomb (C),

and e has been determined experimentally to have the value 1.60×10−19 C.

Charges of larger magnitude than the charge on an electron or on a proton are built up on an object

by adding or removing electrons. Thus, any charge of magnitude q is an integer multiple of e; that

is, q = Ne, where N is an integer.

Example

How many electrons are there in one coulomb of negative charge?

q = 1 C, e = 1.60×10−19 C

q = Ne → 18

191025.6

1060.1

1

e

qN electrons

1.1 Coulomb’s Law

Coulomb’s law states that the magnitude F of the electrostatic force exerted by one point charge

q1 on another point charge q2 is directly proportional to the magnitudes 1q and 2q of the charges

and inversely proportional to the square of the distance r between them:

2

21

r

qqF

Mathematically,

2

21

r

qkqF ,

04

1

k

2

0

21

4 r

qqF

where F : magnitude of electrostatic force

k : electrostatic (Coulomb) constant = 9.0 × 109 N m2 C-2

q1, q2 : magnitude of charge (without sign)

r : distance between two point charges

ε0 : permittivity of free space = 8.85× 10-12 C2 N-1 m-2

Notes:

The sign of the charge can be ignored when substituting into the Coulomb’s law equation.

The sign of the charges is important in distinguishing the

direction of the electric force.



Equation above gives only the magnitude of the electrostatic force that each point charge exerts on

the other; it does not give the direction. The electrostatic force is directed along the line joining the

charges, and the force is attractive if the charges have unlike signs and repulsive if the charges have like signs:

Unlike charge: ATTARCTIVE Like charge: REPULSIVE

The two forces obey Newton’s third law; they are always equal in magnitude and opposite in

direction.

Example: 2 charges

Two objects, whose charges are +1.0 and −1.0 C, are separated by 1.0 km. Compared to 1.0 km, the

sizes of the objects are small. Find the magnitude of the attractive force that either charge exerts on

the other.

N 100.9

101

0.10.1109 3

23

9

2

21

r

qkqF

Example: 3 charges

Determine the magnitude and direction of the net electrostatic force on q1.

Step 1: Sketch force diagram

Step 2: Apply Coulomb’s law equation

N 7.220.0

1041031092

669

2

12

2112

r

qkqF ;

N 4.815.0

1071031092

669

2

13

31

13

r

qkqF

Step 3: Add electric forces as vector

N 7.54.87.21312

FFF

The magnitude of the net electrostatic force acting on q1 is 5.7 N, and it points to the right.

Ignore negative sign of the charge

Unlike charge:

Attract

Unlike charge:

Attract

To the left (−ve) To the right (+ve)

(to the right)



1.2 Electric Field

The electric field

E that exists at a point is the electrostatic force

F experienced by a small test

charge q0 placed at that point divided by the charge itself.

0q

FE

The electric field is a vector, and its direction is the same as the direction of the force on a positive

test charge, or opposite to the direction of the force on a negative test charge. Test charge (q0) is

the charge whose magnitude is very small, in fact negligible, as compared to that of the point charge,

and which does not affect the electric field of the point charge. To find out whether there is an electric

field at a particular point, test charged is placed at that point. If the test charge experiences an electric

force, then there is an electric field at that point.

SI unit of electric field: newton per coulomb (N C-1).

Positive Point Charge

A positive test charge A negative test charge

Negative Point Charge

A positive test charge A negative test charge

*The direction of electric field strength, E depends on sign of isolated point charge.

* The direction of the electric force, F depends on the sign of isolated point charge and test charge.

Example

Given an electric field has a magnitude of 2.0 N C-1 and is directed to the right. Determine the force

on a charge placed at that spot, if the charge has a value of

(a) q0 = +18×10-8 C and

(b) q0 = −24×10-8 C.

a) N 10360.21018 88

0 EqF

Since q0 is positive, the force points in the same direction as the electric field, which is to the right.

b) N 10480.21024 88

0 EqF

Since q0 is negative, the force points in the direction opposite to the electric field, which is to the

left.



Since

0q

FE

and 2

0

r

kQqF

The electric field for a point charge can be calculated by using

2r

kQE

where E : magnitude of electric field strength (intensity)

Q : magnitude of isolated point charge (without sign)

r : distance between a point and isolated point charge

Electric field around charges can be represented by drawing a series of lines. These lines are called

electric field lines (lines of force). The direction of electric field is tangent to the electric field line at

each point. The shape of electric field is sphere.

Isolated point charge

Single positive charge Single negative charge

The lines point radially outward from the

charge

The lines point radially inward from the

charge

Two equal point charges

Unlike charges, +Q and −Q Like charges, +Q and +Q

The lines are curved and they are directed

from the positive charge to the negative

charge.

Neutral point, X is defined as a point

(region) where the total electric force is

zero.

It lies along the vertical dash line.

Notes:

0qQ

Q is isolated point charge

q0 is test charge

+ –



Two opposite unequal charges, +2q and –q

Note that twice as many lines leave +2q as

there are lines entering –q

Number of lines is proportional to

magnitude of charge.

Two opposite charged parallel metal plate

The lines go directly from positive plate to

the negative plate.

The field lines are parallel and equally

spaced in the central region far from the

edges but fringe outward near the edges.

Thus, in the central region, the electric field

has the same magnitude at all points.

The fringing of the field near the edges can

be ignored because the separation of the

plates is small compared to their size.

Notes:

The closer the lines, the stronger the field.

Electric field lines start on positive charges and end on negative charges.

The field lines never cross because the electric field doesn’t have two values at the same point.

Example: 2 point charges

Two point charges are separated by a distance of 10.0 cm. One has a charge of −25 µC and the other

+25 µC. Determine the direction and magnitude of the electric field at a point P between the two

charges that is 2.0 cm from the negative charge.

Step 1: Sketch field diagram

−ve charge: Electric

field line points inward

(E1 toward charge Q1)

+ve charge: Electric field

line points outward

(E2 away from charge Q2)



Step 2: Apply electric field equation

1-8

2

69

2

1

1

1 C N 1063.502.0

1025109

r

QkE

1-7

2

69

2

2

2

2 C N 1003.708.0

1050109

r

QkE

Step 3: Add electric fields as vector

1-878

21 C N 1033.61003.71063.5

EEE (to the left)

The magnitude of the net electric at point P is 6.33×108 N C-1, and it points to the left.

1.3 Electric Potential

Electric potential, V is defined as the work done in bringing positive test charge from infinity to

that point in the electric field per unit test charge, q0.

0q

WV

where W∞ : work done in bringing positive test charge from infinity to that point

q0 : value of test charge (including sign)

Electric potential is a scalar quantity.

The SI unit of electric potential is volt (V) or joule per coulomb (J C-1).

The total electric potential at a point in space is equal to the algebraic sum of the constituent

potentials at that point. In other words, when two or more charges are present, the potential

due to all the charges is obtained by adding together the individual potentials :

VTotal at P = VA + VB

The electric potential at infinity is considered zero.

The electric potential can be positive, negative or zero depending on the signs and magnitudes

of W∞.

If the value of work done is negative – work done by the electric force (system).

If the value of work done is positive –work done by the external force or on the system.

In the calculation of W and V, the sign of the charge MUST be substituted in the related equations.



Equipotential Lines and Surfaces

Equipotential lines and surfaces is defined as a surface where all points on the surface that have the

same electric potential.

A point charge A Uniform Electric Field

The dashed lines represent the equipotential surface (line).

The equipotential surfaces (lines) always perpendicular to the electric field lines passing through them and points in the direction of decreasing potential.

CBA VVV

From the figures, then the work done to bring a test charge from B to A is given by

0

BAo

ABoBA

VVq

VqW

Electric Potential Created by a Point Charge

Extra Notes:

r

kQqW

rkQqW

drrkQqW

r

kQqFdrFdW

drFdW

drFdW

r

r

E

rx

x

E

rx

x

E

ext

0

1

0

2

0

2

0 where

No work is done in moving a

charge along the same

equipotential surface.



Since

0q

WV and

r

kQqW 0

Thus, electric potential created by a point charge Q can be written as

r

kQV

where V : electric potential at a point in electric field

Q : value of a point charge (including sign)

r : distance between a point and isolated point charge

Potential Difference

Potential difference between two points in an electric field is defined as the work done in bringing a

positive test charge from a point to another point in the electric field per unit test charge.

Mathematically,

0q

WVVV initialfinal OR

0q

WVVV BA

BAAB

where WBA : work done in bringing positive test charge from point B to point A

VA : total electric potential at point A (final state)

VB : total electric potential at point B (initial state)

q0 : value of test charge (including sign)

The work done to bring a charge from one point to another in the field does not depend on the path

taken (because the work done by conservative force).

Example:

Where WBA is the work done to

bring a charge q from B to A.

Example:

The work done to bring a

charge from B to A along the

paths 1,2 or 3 is the same.



Example:

Calculate the potential difference between point A and B.

Step 1: Calculate total potential at point A

V 240

4.0

108109

2.0

108109 9999

ve

ve

ve

veveveA

r

kQ

r

kQVVV

Step 2: Calculate total potential at point B

V 0

4.0

108109

4.0

108109 9999

ve

ve

ve

veveveB

r

kQ

r

kQVVV

Step 3: Apply potential difference equation

V 2400240 BAAB VVV

Change in Potential Energy

Extra Notes:

The total work done W in bringing the test charge (+q0) from B to A is given by

UUUUW

r

kQqU

rrkQqW

rkQqW

drrkQqW

r

kQqFdrFdW

ABBABA

BA

r

r

r

r

E

Ax

Bx

E

rx

x

A

B

A

B

0

0

1

0

2

0

2

0

, 11

where

Since 0q

WV or

0q

WV BA

AB , thus

VqU 0 OR ABAB VqU 0

where VAB = Vfinal – Vinitial

UAB = Ufinal – Uinitial

Must include the sign



Potential Energy of a System of Point Charges

The total electric potential energy of the system of charges is the total work done to bring all the charges from infinity to their final positions.

In the system of charges, suppose there were originally no charges at the points A, B and C as in the

figure above.

Step 1:

The charge Q1 is brought from infinity and placed at point A. Since originally there were no charges,

the charge Q1 does not experience any electric force when it is brought from infinity, that is F = 0.

No work is done to place the charge at point A. Hence VA = 0. Since U1 = Q1VA, hence

U1 = 0

Step 2:

With the charge Q1 fixed at point A, an electric field is produced by the Q1. Work is needed to place

Q2 at point B. Hence, the electric potential to bring the charge Q2 from infinity to the point B at a

distance of r12 from A is

12

1

r

kQVB

Since U2 = Q2VB, hence

12

212

r

QkQU

Step 3:

With the charge Q1 fixed at point A and Q2 at the point B, the electric potential at the point C is

23

2

13

1

r

kQ

r

kQVC

Since U3 = Q3VC, hence

23

32

13

313

r

QkQ

r

QkQU

The total potential energy, U can be expressed as

23

32

13

31

12

21

23

32

13

31

12

21

321

0

r

QQ

r

QQ

r

QQkU

r

QkQ

r

QkQ

r

QkQU

UUUU

A

B

C

A system of TWO charges

A system of THREE charges



1.4 Charge in a Uniform Electric Field

Case 1: Stationary charge

WFe → Static Equilibrium

Positive Charge Negative Charge

Consider a particle of charge q0 and mass m is stationary and placed in a uniform electric field

E, the electric force Fe exerted on the charge is given by

EqFe 0

Since the particle is stationary the net force acting on the particle is zero. So the electric force

on the particle, Fe is equal to the weight, W of the particle.

mgEq

WF

WF

F

e

e

0

0

0

WFe

If WFe , the charged particle will be attracted toward the direction with greater force.

WFe WFe

mamgEq

maWF

maF

e

0

maEqmg

maFW

maF

e

0

WFe → W = 0 and EqFe 0

If the mass of the charged particle is negligible, the particle will be attracted toward the plate

with opposite charge.




Case 2: Charge moving perpendicularly to the field

Consider a particle of charge q0 and mass m enters a uniform electric field, E perpendicularly

with an initial velocity u, the upward electric force will cause the charged particle to move

along a parabolic path towards the upper plate.

From Newton’s second law

maF

The electric force Fe exerted on the charge is

EqFe 0

Since the path makes by the electron is similar to the motion of a ball projected horizontally

above the ground,

0xa , uux , 0yu

Therefore the magnitude of the electron’s acceleration is given by

m

Eqaa o

y

The components of electron’s velocity after pass through the electric field are given by

uuv xx ; tauv yyy

tm

Eqvy 0

The position of the electron is

tus xx ; 2

2

1tatus yyy

20 2

1t

m

Eqs y

Notes: Charged particle is moving in a straight

path before it enters the plate and after it leaves

the plate.

Straight

path

Straight

path Parabolic

path



Case 3: Charge moving parallel to the field


If the electric force on the positive charge is

in the same direction as to its motion, the

positive charge accelerates along a straight

line.

If the electric force on the positive charge is

in the opposite direction to its motion, the

positive charge decelerates along a straight line.

If the electric force on the negative charge

(electron) is in the same direction as to its

motion, the positive charge accelerates

along a straight line.

If the electric force on the negative charge

(electron) is in the opposite direction to its

motion, the positive charge decelerates

along a straight line

Consider a particle of charge q0 and mass m is moving parallel to the uniform electric field E,

the electric force Fe exerted on the charge is given by

EqFe 0

Since only electric force Fe exerted on the particle, thus this force contributes the net force, F and causes the particle to accelerate.

According to Newton’s second law,

m

Eqa

maEq

maF

maF

e

0

0

Because the electric field is uniform (constant in magnitude and direction) then the acceleration of the particle is constant.



Case 4: Charge in dynamic equilibrium

Dynamic equilibrium means the charge moves with constant velocity perpendicularly to the

direction of electric field and the total force acting on charge is zero, 0F .

If the particle travels in a straight line with constant velocity hence the electric and its weight

are equal in magnitude.

mgEq

WF

WF

F

e

e

0

0

0

Only particles with this constant speed can pass through without being deflected by the fields.

If WFe , the charged particle will be reflected toward the direction with greater force.

Uniform Electric Field (eg. between Two Parallel Plates)

Consider a uniform electric field is produced by a pair of flat metal plates, one at which is earthed

and the other is at a potential of +V as shown in figure.

As the positive test charge travel from A to B :

0q

WV and EdqsFW 0

EdV

d

VE

A B

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Chapter 1: Electrostatics - YSL...

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