Electrostaticsp. 213

Negative Charge - Addition (gain) of e-

+ +

+ +-

-

-

-

+ +

+ +-

-

-

-

+ +

+ +-

--

-

-

-

-

-

Positive Charge - Removal (loss) of e-

+ +

+ +-

-

+ +

+ +-

-+ +

+ +-

- ---

-

Similar charges repel each other.

-- +

-- -- --

Unlike charges attract each other.

+++

-- --- -

Law of Conservation of Charge:

Charge cannot be created or destroyed,

it can only be transferred from one

object to another.

Coulomb’s Law:

The magnitude of the electrostatic force

exerted by two point charges (Q₁ and Q₂)

on each other is directly proportional to the

product of the magnitudes of the charges

and inversely proportional to the square of

the distance (r) between them.

2

21

r

QkQ F =

F – Electrostatic force in Newton

Q1 and Q2 – Charges in Coulomb

r – Distance between charges in m

k – proportionality constant

(9 x 109 N.m2.C-2)

Coulomb’s Law

(Electrostatic forces between charges)

Two factors influence

the force:

• The magnitude of the

charges.

• The magnitude of the

distance between the

charges.

21QQ α F

2r

1 α F

ExampleThe centres of two identical spheres are 6 mm

apart. They carry charges of 5 nC and -10 nC

respectively as shown in the diagram.

1. Calculate the magnitude of the electrostatic

force between the two spheres.

Q1 Q2

5 nC

-10 nC

6mm

F = 𝑘𝑄1𝑄2

𝑟2

= (9×109)(5×10−9)(10×10−9)

(0,006)2

= 1,25 × 10-2 N, attractive

It is not necessary to substitute the signs of the charges.

Example

The centres of two identical spheres are 6 mm

apart. They carry charges of 5 nC and -10 nC

respectively as shown in the diagram.

2. Is it a force of attraction or repulsion?

Q1 Q2

5 nC

-10 nC

6mm

Attraction

Example

The centres of two identical spheres are 6 mm

apart. They carry charges of 5 nC and -10 nC

respectively as shown in the diagram.

3. The spheres touch each other, and are then

moved back to their original positions. What

charge does each sphere have now?

Q1 Q2

6mm

Qon each after separation = 𝑄1+𝑄2

2

= +5×10−9 + (−10×10−9)

2

= -2,5 x 10-9 C

Example

The centres of two identical spheres are 6 mm

apart. They carry charges of 5 nC and -10 nC

respectively as shown in the diagram.

4. How many electrons are transferred when

they touch, and in which direction?

Q1 Q2

6mm

ΔQ1 = Q1new - Q1original

= (-2,5x10-9) - (+5x10-9)

= -7,5 x 10-9 C

Thus, number of electrons transferred:

ne- = 7,5×10−9

1,6×10−19

= 4,7 x 1010 electrons, from Q₂ to Q₁

Homework

p. 224,

nos. 3, 6.1, 6.2, 11, 13, 14

Electrons are removed

Forces in two dimensions:

BC

A -

++F

A o

n C

FB on C Cθ

Additional examples:1. For the charge configuration shown, calculate the charge on Q₃ if the resultant force on

Q₂ is 6,3 x 10-5 N to the right, and Q1 = 4,36 x 10-6 C, Q2 = -7 x 10-7 C, r1 = 1,85 x 10-1 m,

r2 = 4,7 x 10-2 m.

Q3Q1 Q2

r1 r2

FQ1 on Q2 = k𝑄1𝑄2

𝑟2 = 9 x 109 x (4,36×10−6)×(7×10−7)

(1,85×10−1 + 4,7×10−2)2

= 0,51 N, to the left

+ ? -

Fnet = FQ1 on Q2 + FQ3 on Q2

6,3 x 10-5 = -0,51 + FQ3 on Q2

FQ3 on Q2 = 5,1 x 10-1 N, to the right

Additional examples:1. For the charge configuration shown, calculate the charge on Q₃ if the resultant force on

Q₂ is 6,3 x 10-5 N to the right, and Q₁ = 4,36 x 10-6 C, Q₂ = -7 x 10-7 C, r₁ = 1,85 x 10-1 m,

r₂ = 4,7 x 10-2 m.

FQ3 on Q2 = k𝑄3𝑄2

𝑟2 = 9 x 109 x ( 𝑄3 )×(7×10−7)

(4,7×10−2)2

5,1 x 10-1 = 2 851 969,217 x Q3

∴ Q3 = -1,79 x 10-7 C

Q3Q1 Q2

r1 r2

+ - -

Additional examples:2. Calculate the resultant force on Q₁ given the following charge configuration:

Q2Q1

Q3

0,04m

+3nC

FQ3 on Q1 = k𝑄1𝑄2

𝑟2 = 9 x 109 x (3×10−9)×(2×10−9)

(0,04)2

= 3,375 x 10-5 N, downwards

0,07m

+2nC

+1nC

Additional examples:2. Calculate the resultant force on Q₁ given the following charge configuration:

Q2Q1

Q3

0,04m

+3nC

FQ2 on Q1 = k𝑄1𝑄2

𝑟2 = 9 x 109 x (1×10−9)×(2×10−9)

(0,07)2

= 3,673 x 10-6 N, to the left

0,07m

+2nC

+1nC

Additional examples:2. Calculate the resultant force on Q₁ given the following charge configuration:

Q1

3,67 x 10-6 N

3,38 x 10-5 N

3,38 x 10-5 N

θ

(Fnet)2 = (3,67 x 10-6)2 + (3,38 x 10-5)2

Fnet = 3,4 x 10-5 N

tan θ = (3,38 x 10-5) ÷ (3,67 x 10-6)

θ = 83,8o

∴ Fnet = 3,4 x 10-5 N; 186,2o from the line connecting Q₃ with Q₁

Additional examples:3. Calculate the resultant force on Q₁ given the following charge configuration:

Q2Q1

Q3

0,4m

-3nC

FQ3 on Q1 = k𝑄1𝑄2

𝑟2 = 9 x 109 x (3×10−9)×(9×10−9)

(0,4)2

= 1,52 x 10-6 N, downwards

0,65m

-9nC

+1nC

Additional examples:3. Calculate the resultant force on Q₁ given the following charge configuration:

Q2Q1

Q3

0,4m

-3nC

FQ2 on Q1 = k𝑄1𝑄2

𝑟2 = 9 x 109 x (1×10−9)×(9×10−9)

(0,65)2

= 1,91 x 10-7 N, to the right

0,65m

-9nC

+1nC

Additional examples:3. Calculate the resultant force on Q₁ given the following charge configuration:

Q11,91 x 10-7 N

1,52 x 10-6 N

1,52 x 10-6 N

θ

(Fnet)2 = (1,52 x 10-6)2 + (1,91 x 10-7)2

Fnet = 1,53 x 10-6 N

tan θ = (1,52 x 10-6) ÷ (1,91 x 10-7)

θ = 82,8o

∴ Fnet = 1,53 x 10-6 N; 172,8o from the line connecting Q₃ with Q₁

Additional examples:4. Calculate the resultant force on Q₂ given the following charge configuration:

Q2Q1

Q3

0,05m

+8nC

FQ1 on Q2 = k𝑄1𝑄2

𝑟2 = 9 x 109 x (8×10−9)×(3×10−9)

(0,05)2

= 8,64 x 10-5 N, to the right

0,03m

-2nC

+3nC

Additional examples:4. Calculate the resultant force on Q2 given the following charge configuration:

FQ3 on Q2 = k𝑄1𝑄2

𝑟2 = 9 x 109 x (2×10−9)×(3×10−9)

(0,03)2

= 6 x 10-5 N, downwards

Q2Q1

Q3

0,05m

+8nC

0,03m

-2nC

+3nC

Additional examples:4. Calculate the resultant force on Q2 given the following charge configuration:

Q28,64 x 10-5 N

6 x 10-5 N

6 x 10-5 N

θ

(Fnet)2 = (8,64 x 10-5)2 + (6 x 10-5)2

Fnet = 1,05 x 10-4 N

tan θ = (6 x 10-5) ÷ (8,64 x 10-5)

θ = 34,78o

∴ Fnet = 1,05 x 10-4 N; 124,8o from the line connecting Q₂ with Q₃

Homework:

p. 224

nos. 20, 23 (change the mass to 0,1g), 28

Electric field

An electric field as a region in

space in which an electric

charge experiences a force.

Electric fields

The shape of an electric field is

determined by the charge that

causes it.

The direction of the electric field at a point is

the direction in which a positive test charge

would move if placed at that point.

Properties of electric field lines

• Start and end perpendicular to the surface

of a charged object.

• May never cross.

• Field lines close together – strong field;

field lines further apart – weaker field.

• Direction by convention.

• Surround charged object in 3 dimensions.

Uniform electric field (not syllabus)

Exists between two parallel plates which

are oppositely charged.

A small positive test charge experiences

the same electrostatic force at any point

in this electric field.

+

-

Point charges

There is no uniform electric field around a point charge.

The further away from the point charge, the weaker the

electrostatic force.

There is no charge or electric field inside a hollow

charged object.

+ -

Charges with irregular shapes (not syllabus)

The strongest field is around the sharpest point.

+

Electric field strength

q

F E =

E = Electric field strength in N.C-1

F = Force in N

q = Charge in C (test charge)

Electric field strength at a POINT

in an electric field is the force per

positive unit charge at that point.

Electric field strength

Electric field strength around

a point charge or charged object.

Field strength around a charge “Q” at

a distance “r” from charge “Q”.

2r

kQ E =

2r

1 α Een

Q α E

q

F E =

2r

kQ E =

q experiences

field strength

Q causes

field strength

Homework

p. 247

nos. 10.1, 12.4, 13, 18.1

, away from the 5μC