chapter 1: electrostatic - ysl physics -...

Chapter 1

BP3 1 FYSL

Chapter 1: Electrostatic

1.1 Coulomb’s Law

L.O 1.1.1 State Coulomb’s law

Coulomb’s law states that the magnitude of the electrostatic (Coulomb/electric) 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.

Mathematically,

The S.I. unit of charge is coulomb (C).

LO 1.1.2 Sketch the electric force diagram and apply Coulomb’s law for a system

of point charges. (2D, maximum four charges)

Like charges – REPEL Unlike charges – ATTRACT

Example:

Sketch the force diagram for q1.

Sketch the force diagram for q1.

2

4 2

0 r

Qqk

r

QqF

Electrostatic constant

(9.0 × 109

N m2

C-2

)

Permittivity of free space

(8.85× 10-12

C2

N-1

m-2

)

where

Q,q : magnitude of charges

r : distance between two point charges

+ –

+

+ – q1

q1

q2

q3

–

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.

2r

QqF

Chapter 1

BP3 2 FYSL

Example

Question Solution

Two point charges, q1 = −20 nC and q2 = 90 nC, are

separated by a distance of 4.0 cm as shown in

figure below.

Find the magnitude and direction of

a. the electric force that q1exerts on q2.

b. the electric force that q2 exerts on q1.

(Given Coulomb’s constant, k = 9.0×109 N m

2 C

-2)

Three point charges lie along the x-axis as shown in

figure below.

Calculate the magnitude and direction of the total

electric force exerted on q2.


2 C

-2)

Figure below shows the three point charges are

placed in the shape of triangular.

Determine the magnitude and direction of the

resultant electric force exerted on q1.


2 C

-2)

– +

q1 q2 4.0 cm

q1 = 2C q2 = 4C q3 = 6C

3.0 cm 5.0 cm

+ + –

–

– +

q1 = 1.2 µC q2 = 3.7 µC

q3 = 2.3 µC

15 cm

10 cm

32°

Chapter 1

BP3 3 FYSL

Example

Question Solution

Two point charges, q1 = +4.0µC and q2 = + 6.0µC,

are separated by a distance of 50 cm as shown in

figure below.

Determine the position of a point charge q that is

placed on the line joining q1 and q2 such that the net

force acting on it is zero.


2 C

-2)

Exercise

Question

The electron in the hydrogen atom orbits around the proton in an orbit of radius 5.3×10-11

m.

Determine the force between the electron and the proton in a hydrogen atom.

(Given Charge of proton = +1.60×10-19

C, Charge of electron = − 1.60×10-19

C)

Answer: − 8.2×10-8

N

Point charges Q1 = +10 µC, Q2 = − 15 µC and Q3 = + 5.0 µC are placed at the corners of a

triangle as shown in figure below. Determine the resultant force that acts on charge Q3.

Answer: 0.037 N

Four identical point charges (q = +10.0 C) are located on the corners of a rectangle as

shown in figure below.

Calculate the magnitude and direction of the resultant electric force exerted on the charge at

the lower left corner by the other three charges. Answer: 40.9 N, 83° below –ve x-axis)

+ +

q1 q2 50 cm

Q1

Q2 Q3

3 m

4 m +

+

–

+

+

+

+

60 cm

15 cm

Chapter 1

BP3 4 FYSL

1.2 Electric Field

LO 1.2.1 Define and Use Electric Field Strength, oq

FE

LO 1.2.2 Use 2r

kQE for a point charge

Electric field strength (intensity) is defined as the electric (electrostatic) force per unit

positive test charge that acts at that point.

For a point charge,

The S.I. unit is N C1

OR V m 1

.

LO 1.2.3 Sketch the electric field strength diagram and determine electric field

strength E for a system of charges. (2D, maximum 4 charges)

Electric field is defined as a region of space around isolated charge where an electric

force is experienced if a (positive) test charge is placed in the region.

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

where

F : magnitude of the electric force

qo : magnitude of test charge

22 4 r

Q

r

kQE

o

where

Q : magnitude of the isolated charge

r : distance between the point and isolated

point charge

+ –

oq

FE

Notes: In the calculation of magnitude E, substitute the magnitude of the charge only.

Chapter 1

BP3 5 FYSL

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.

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.

Chapter 1

BP3 6 FYSL

Example

– – – –

− Q − Q − Q − 3Q

Note!!! • 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.

Since both F and E are vectors, how

to determine the direction of F and E

for a 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.

Chapter 1

BP3 7 FYSL

Example

Question Solution

A metal sphere can be regarded as a point object in

space. It carries a charge of +6.0 µC. Find the

electric field that the sphere generates at a distance

of 10 cm around it.

Two point charges, q1= − 1 C and q2= + 4 C, are

placed 2 cm and 3 cm from the point A respectively

as shown in figure below.

Find

a. the magnitude and direction of the electric field

intensity at point A.

b. the total electric force exerted on q0=-4 C if it

is placed at point A.


2 C

-2)

Two point charges, q1= − 12nC and q2= 12nC, are

placed 0.10 m apart. Calculate the total electric

field at point a and point b.

– +

q1 q2 2 cm 3 cm

A

6 cm

– +

q1 q2 4 cm

b

a

13 cm 13 cm

Chapter 1

BP3 8 FYSL

Exercise

Question

Two point charges of +2.0 µC and −4.0 µC are separated by a distance of 5.0 cm. Determine

the electric field at the midpoint between two charges.

Point charges of +2.0 µC and −2.0µC are placed at corners A and B respectively of a right-

angle triangle ABC, as shown in the figure below. Determine the resultant electric field

strength at corner C. Answer: 6.85 × 106 N C

-1 , 81.79° above +vex-axis

Find the magnitude and direction of the electric field at the centre of the square in figure

below if q = 1.0×10-8

C and a = 5 cm. Answer: 1.02×105 N C

-1, upwards.

– +

A B

C

12.0 cm

5.0 cm

Chapter 1

BP3 9 FYSL

1.3 Electric Potential

LO 1.3.1& 1.33: Define Electric Potential & Use r

kQV

Electric potential, V of a point in the electric field is defined as the work done in bringing

positive test charge from infinity to that point in the electric field per unit test charge.

OR Electric potential, V of a point in an electric field is defined as the potential energy per

unit positive charge at that point in the electric field.

The S.I. unit is V or J C-1

.

Since r

kQqU o , then the equation of electric potential can be written as

Note!!

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

constituent potentials at that point. BA VVV Total

The theoretical zero of electric potential of a charge is at infinity.

The electric potential energy of a positively charged particle increases

when it moves to a point of higher potential.

The electric potential energy of a negatively charged particle increases

when it moves to a point of lower potential.

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

magnitudes of qo and 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 U, W and V, the sign of the charge MUST be substituted in the

related equations.

where

W∞ : work done

qo : magnitude of test charge oq

WV

oq

UV

where

U : potential energy

qo : magnitude of test charge

VqU o

r

kQV

Chapter 1

BP3 10 FYSL

LO 1.3.2: Define and sketch equipotential lines and surfaces of an isolated charge

and a uniform electric field.

Equipotential surface (line)is defined as the locus of points 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

LO 1.3.4: Calculate potential difference between two points

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,

OR

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).

No work is done in moving a

charge along the same

equipotential surface.

oq

WVVV initialfinal

o

BABAAB

q

WVVV

Chapter 1

BP3 11 FYSL

LO 1.3.5: Deduce the change in potential energy between two points in electric field.

From the definition of electric potential difference,

oq

WV and UW

LO 1.3.6: Calculate 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.

The total potential energy, U can be expressed as

23

32

13

31

12

21

23

32

13

31

12

21

231312

r

QQ

r

QQ

r

QQkU

r

QkQ

r

QkQ

r

QkQU

UUUU

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.

VqU o

Chapter 1

BP3 12 FYSL

Example

Question Solution

Figure below shows a point A at distance 10 m

from the positive point charge, q = 5C.

Calculate the electric potential at point A and

describe the meaning of the answer.


2 C

-2)

Two point charges, q1=+0.3 C and q2= −0.4 C are

separated by a distance of 6 m as shown in figure

below. Calculate the electric potential at point A if

pointA is at the midpoint of q1 and q2.


2 C

-2)

Two point charges, q1=+12 nC and q2= −12 nC are

separated by a distance of 8 cm as shown in figure

below.

Determine the electric potential at point P.


2 C

-2)

+

10 m

q A

+

6.0 m

q1

A −

+

8.0 cm

P

−

q2

q1 q2

6.0 cm

Chapter 1

BP3 13 FYSL

Example

Question Solution

Two points, S and T are located around a point

charge of +5.4 nC as shown in figure below.

Calculate

a. the electric potential difference between points

S and T.

b. the work done in bringing a charge of 1.5 nC

from point T to point S.


2 C

-2)

Two point charges, Q1= +2.0 C and Q2= 6.0 C,

are placed 4.0 cm and 5.0 cm from a point

respectively as shown in figure below.

a. Calculate the electric potential at P due to the

charges.

b. If a charge Q3= +3.0 C moves from infinity to

P, determine the change in electric potential

energy for this charge.

c. When the charge Q3 at point P, calculate the

electric potential energy for the system of

charges.


2 C

-2)

+

6.0 cm

S 8.0 cm T

+

5.0 cm

4.0 cm P

−

Q1

Q2

Chapter 1

BP3 14 FYSL

Exercise

Question Solution

Two point charges, Q1= 40 C and Q2= 30 C

are separated by a distance of 15 cm as shown in

figure below.

Calculate

a. the electric potential at point A and describe

the meaning of the answer,

b. the electric potential at point B.


2 C

-2)

Answer: VA = −9.89×106 V , VB = −3.88×10

6 V

Two point charges q1=+2.40 nC and q2= −6.50 nC

are 0.1 m apart. Point A is midway between them;

point B is 8 cm from q1 and 6cm from q2 as shown

in figure below.

Find

a. the electric potential at point A,

b. the electric potential at point B,

c. the work done by the electric field on a charge

of 2.5 nC that travels from point B to point A.


2 C

-2)

Answer: VA = −738 V , VB = −705 ,

WBA = −8.25×10-8

J

− 5 cm

13 cm

B

− A 10 cm

5 cm

+ −

q1 q2 5 cm

B

A

6 cm 8 cm

Q1 Q2

Chapter 1

BP3 15 FYSL

1.4 Charge In A Uniform Electric Field

LO 1.4.1: Explain quantitatively with the aid of a diagram the motion of charge in a

uniform electric field.

Case 1 : Stationary Charge

Positive stationary charge Negative stationary charge

Force experienced by charge is in the same

direction as electric field, E.

Force experienced by charge is in the

opposite direction as electric field, E.

• Consider a stationary particle of charge qo and mass m is placed in a uniform electric

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

EqF oe

• Since only electric force exerted on the particle, thus this force contributes the nett force,

F and causes the particle to accelerate.

• According to Newton’s second law, then the magnitude of the acceleration of the

particle is

maEq

maFF

o

e

m

Eqa e

• Because the electric field is uniform (constant in magnitude and direction) then the

acceleration of the particle is constant.

• If the particle has a positive charge, its acceleration is in the direction of the electric

field. If the particle has a negative charge (electron), its acceleration is in the direction

opposite the electric field.

− − − − −

+ + + + +

− − − − −

+ + + + +

+

−

F

F

Chapter 1

BP3 16 FYSL

Case 2 : Charge moving perpendicularly to the field

Positive charge Negative charge

• The positive charge will be deflected and

moves along a parabolic path towards

the negative plate.

• The positive charge moves under the

influence of the electric force which is at

the same direction as electric field

lines.

• The negative charge will be deflected and

moves along a parabolic path towards

the positive plate.

• The negative charge moves under the

influence of the electric force which is

opposite direction to the electric field

lines.

• Consider an electron (e) with mass, me enters a uniform electric field, E perpendicularly

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

along a parabolic path towards the upper plate.

• From Newton’s second law,

maF

• The electric force Fe exerted on the charge,

EqF oe

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

upward) :(direction m

eEa y since 0xa

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

above the ground.

x - component y – component

u

v

s

a

− − − − −

+ + + + +

− − − − −

+ + + + +

+

− +

−

v

v

Chapter 1

BP3 17 FYSL

Case 3 : Charge moving parallel to the field

Fe and v → in the same direction Fe and v → in the opposite direction

• The electric force on the positive charge

is in the same direction as to its motion.

• The positive charge accelerates along a

straight line.

• The electric force on the positive charge

is in the opposite direction to its motion.

• The positive charge decelerates along a

straight line.

Case 4 : Charge in dynamic equilibrium

Between electric force and weight

Particle weight is at the opposite

direction to the electric force.

WFe

Dynamic equilibrium means the charge moves with constant velocity.

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

fields.

− − − − −

+ + + + +

+

v

v

F

F

+

− − − − −

+ + + + +

F

W

−

− − − − −

+ + + + +

− −

Chapter 1

BP3 18 FYSL

LO 1.4.2: Use d

VE

for uniform electric field.

The graph is a straight line with negative constant gradient, thus

d

VE

d

V

r

VE

0

0

Example

Question Solution

Two parallel plates are separated 5.0 mm apart. The

electric field strength between the plates is 1.0

104 N C

1. A small charge of +4.0 nC is moved

from one conducting plate to another. Calculate

a. the potential difference between the plates

b. the work done on the charge.

Chapter 1

BP3 19 FYSL

Question Solution

If the plates are horizontal and separated by 1.0 cm,

the plates are connected to 100 V battery, the

magnitude of the electric field is 1.0×104 N C

-1. If

an electron is released from the rest at the upper

plate, determine

a. the acceleration of the electron.

b. the speed and the kinetic energy required to

travel to the lower plate.

c. the time required to travel to the lower plate.

Question Solution

The figure shows a section of the deflection system

of a cathode ray oscilloscope. An electron

travelling at a speed of 1.5×107 m s

-1 enters the

space between two parallel metal plates 60 mm

long. The electric field between the plates is

4.0×103 V m

-1.

a. Copy the figure, sketch the path of the electron

in between plates, and after emerging from the

space between the plates.

b. Find the acceleration of the electron between

the plates.

c. Determine the velocity when it emerges from

the space between the plates.

20 mm

60 mm

80 V

0 V

1.5×107 m s

-1

Chapter 1

BP3 20 FYSL

Exercise

Question Solution

Two parallel electrodes X and Y are separated by a

distance of 1.80 cm. The electric field between the

electrodes is 2.4 x 104 NC

-1. Electrons are emitted

with negligible velocity from the electrode X.

Determine

a. the force in the electric field

b. the acceleration of the electron in the electric

field.

c. the velocity of the electron when it reaches the

electrode Y.

d. the time taken by the electron to travel from

the electrode X to Y.

(Given e=1.60 1019

C and me=9.11 1031

kg)

Answer: 3.84 x 10-15

N, 4.22 x 1015

ms-2

,

1.23 x 107 ms

-1, 2.93 x 10

-9 s

Question Solution

Figure shows an electron entering charged parallel

plates with a speed of 5.45 106 m s

1. The electric

field produces by the parallel plates has deflected

the electron downward by a distance of 0.618 cm at

the point where the electron exits. Determine

a. the magnitude of the electric field,

b. the speed of the electron when it exits the

parallel plates.

(Given e=1.60 1019

C and me=9.11 1031

kg)

Answer: 4126 N C-1

, 6.22×106 m s

-1

chapter 1: electrostatic - ysl physics -...

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