9/1/2024

Electricity & Magnetism – Lecture 1

Electric Charge and Coulomb’s law

Electric Charge
 It is the basic property of matter

 There are two kinds of charges: Positive and Negative charges + -

Ebonite rod & Fur  Negatively charged ebonite rod

+ SILK
+ - - -
+
Glass rod & Silk  Positively charged glass rod
+ - -
+
Glass Rod

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Electric Charge
Atom:
Nucleus (small, massive, positive charge)
Electron cloud (large, very low density, negative
charge)

Atom is electrically neutral because of same number of

proton and electron.
Rubbing charges objects by moving electrons from one
to the other.

-
Nucleus
-

-
n + n -
+ + n
+
n n
- +
+ n

- -

Negative
Neutral Atom
Positive Atom
Atom
Number
Numberof
Number ofelectrons
of electrons><=Number
electrons Numberof
Number ofprotons
of protons
protons
-2e -19C
+2e==-3.2
+3.2x x10
10-19 C

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What is Electric Charge?

Excess of electron or deficiency of electron is the charge
Positive Charge: A shortage of electrons.
Negative Charge: An excess of electrons.

 Conservation of charge – The net charge of a closed system remains constant.

 Charge is quantized – integer multiples of the elementary

charge e
q = ne n= ±1, ±2, ±3, ….
e = 1.6 x 10 -19 C

like charges repel and opposite charges attract

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Conductors and Insulators

Substances that readily conduct electric charge are called electrical conductors. Conductors have free
electrons, which conduct the electricity.
Examples: Metals such as copper, aluminum, silver, gold, and tap water.

When atoms of a conductor like copper come together to form the solid, some of their outermost (and
so most loosely held) electrons become free to wander about within the solid, leaving behind
positively charged atoms (positive ions).

We call the mobile electrons conduction or free electrons. There are few (if any) free electrons in a
nonconductor (insulator).

Materials that conduct electric charge poorly are known as electrical insulators.
Examples: Rubber, plastics, dry-wood, and chemically pure water.

Semiconductors are materials that are intermediate between conductors and insulators; examples
include silicon and germanium in computer chips.

Superconductors are materials that are perfect conductors, allowing charge to move without any
hindrance.

Charging by friction:
Static charge
An object can be charged :
1. By friction
2. By conduction
3. By induction

Charging by Contact:

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Charging by Induction

Static electricity is much more in winter

Dry air is a relatively good electrical insulator than moist air

Charging By Friction Charging By Induction

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Coulomb's Law Two charged particles attract each other if they have the
opposite sign of charge, and repel each other if they have
the same signs of charge.
This force of repulsion or attraction due to the charge
properties of objects is called an electrostatic force.
The equation giving the force for charged particles is
called Coulomb's law after Charles-Augustin de
Coulomb.

q1 q 2
F =k .
r2

k = Coulomb’s constant ε0 = permittivity of free space

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Example 1
Two charges are separated by a distance r and have a force F on each other.
q1q2
F =k
r2
F q2 F
q1
r

If r is doubled then F is : ¼ of F

If q1 is doubled then F is : 2F

If q1 and q2 are doubled and r is halved then F is : 16F

Example 2
Two 40 gram masses each with a charge of 3μC are placed 50cm apart. Compare the
gravitational force between the two masses to the electric force between the two masses.
(Ignore the force of the earth on the two masses)

3μC 3μC
40g 40g

50cm

m1 m 2 (. 04 )(. 04 )
Fg = G = 6 .67 × 10 −11 ≈ 4 . 27 × 10 − 13 N
r2 ( 0 .5 ) 2

−6
q1 q 2 9 ( 3 × 10 )( 3 × 10 − 6 )
FE = k 2 = 9 .0 × 10 ≈ 0 .324 N
r ( 0 .5 ) 2
The electric force is much greater than the gravitational force e/m = 1.758820 × 1011 C/kg

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Coulomb’s Law:
The electrostatic force between two charged objects is proportional to the quantity of each of
the charges and inversely proportional to the square of the distance between the charges.
• The strength of the electrostatic force between two charges q1 and q2 is given by Coulomb’s law

  In vector form
  
⃗ =  ̂
 

 9  10 The force on q1 due to q2 is equal and opposite to ⃗ (i.e. -⃗ 

• The direction of the force is along the joining line

The net force

,   +  +  +! +………+ 

Electrostatic force
• Where multiple charges are present, the forces sum as vectors (“principle
of superposition”)
What is the combined force on the blue charge
from the two red charges?

+ve +ve


+ve "#$   % 
+ve

+ve +ve

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Electrostatic force
• Where multiple charges are present, the forces sum as vectors (“principle of
superposition”)

+ve


+ve

+ve

Example 3

Two protons are 3.6 nm apart. What is the total force on an electron
located on the line between them, 1.2 nm from one of the protons?
(elementary charge e1.6 x 10-19 C

q=+e q=-e q=+e

r1=1.2 nm=r r2=2.4 nm=2r

|E ||E |
F
|E ||E |
F

| |   | |  
   (2

F
F 3
F 3  9  10  1.6  10 

(GĴ)
 G   G    0.12 I
 4 4 4  1.2  10 
Direction? Along –ve x-axis direction

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Example 4

Three charges lie on the x axis: q1=+25 nC at the origin, q2= -12 nC at x =2m, q3=+18 nC at x=3 m.
What is the net force on q1? We simply add the two forces keeping track of their directions. Let a
positive force be one in the + x direction.

x
1 2 3

 q2 q3 
F= -kq 1  +
 (2m ) 2 (3m ) 2 
 −12 × 10 −9 C 18 × 10 −9 C 
(
= − 1010 Nm 2
C2 )(25 × 10 −9
C) 
 (2m)2
+
(3m)2 
= 2.5 × 10 −7 N

Direction?

Example 6
Three charged objects are placed as shown. Find the net
force on the object with the charge of -4μC.
q1q 2
F = k
- 5μC r2
45º
(5 ×10−6 )(4 ×10−6 )
2 2 F1 = 9 ×109 = 4.5N
20 + 20 ≈ 28cm (0.20)2
20cm

(5×10−6 )(4×10−6 )
F2 = 9×109 = 2.30N
F1 45º (0.28)2
5μC - 4μC
20cm F2
R= K % L % 2KL MNOP
LMNOP
P  tan ( 
K % LOQIP

F1 and F2 must be added together as vectors.

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R MN S   cos P %  cos P %  cos P

T MN S   sin P %  sin P %  sin P

R R %T


T
P  tan
R

Example 6
Three charged objects are placed as shown. Find the net
force on the object with the charge of -4μC.
q1q 2
F = k
- 5μC r2
45º
(5 ×10−6 )(4 ×10−6 )
2 2 F1 = 9 ×109 = 4.5N
20 + 20 ≈ 28cm (0.20)2
20cm

(5×10−6 )(4×10−6 )
F2 = 9×109 = 2.30N
F1 45º (0.28)2
5μC - 4μC
20cm F2 R MN S   cos P %  cos P
R MN S  4.5 cos 180 % 2.3 cos 315  G2.87
T MN S  4.5 sin 180 % 2.3 sin 315  G1.63

R  R % T = 3.3 N 
T
P  tan  29.4
R

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Example 7 q3= - 2 nC

+y

5 cm
2 cm

Fnet
F13
F12 1 cm +x
q1= + 1 nC q2= + 1 nC

What is the net force on q1 and in what direction?

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