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Coulombs Law:

Subject: Electro Magnetic Field Theory.

Course In charge: Engr. M.Asghar Daudani.

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COULOMB’S LAW

*Q1
*

*Q2
*

**Force due to Q1
acting on Q2
**

**Unit vector in
direction of R12
**

2 120

21 12 4

ˆ
12 *r
*

*QQaF R *

12*r
*

12*F*

2-4 Coulombs Law :

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Coulomb conducted several experiments on charges bodies, he concluded that;

There exists a force between them. He formulated a l aw known as Coulombs law.

The law appears in the following forms; (i). Gaussian’s Form:

**Force; F = ( ) a r .
**

(ii). S I Form:
**Force; F = ( ) ( ) a r
**

(iii). Heavside-Lorentz Form:
**Force; F = ( ) ( ) a r**

Law Statement: Coulombs law states that there exists a force betwee n charged bodies and it is; Proportional to the product of the two charges. Inversely proportional to the square of the distance

between the charges. The force also depends on the medium in which th

e charges are located. The force is a vector quantity and it is attractive if

the charges are unlike and repulsive if the charges are alike.

It acts along the straight line joining the two charg es. 4

Coulombs Law:

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Coulombs Law: Mathematically the coulombs law is given by;

**F ∞ ( ) a r **, Newton Or, **F = **k** ( ) a r
**Where;
k = a constant of proportionality = ( ) m / F-rad.
Q 1 & Q 2 are two point charges (C). (Force Between the

charges in Newton's). r = Distance between the charges (m). ε = Permittivity of medium in which charges are located (F/m) =

**ε 0****ε r** .

**ε 0 = Permittivity of Vacuum or free space =(1/36 Л) x 10) x 10– 9 = 8.854 x 10– 12, F/m.
**

**ε r** = Relative Permittivity of medium w.r.to free space (1 for free space).

** ε r** = ( ) = 9 x 10 9, m / F.
**a** r = Unit vector along the line joining the two charges. 6

Coulombs Law:

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Coulombs Law: The force on Q 2 due to Q 1 in free space is written as in the form of;

**F 2 = ( ) {a 2 1 / (r 2 1) 2}
**

Where;
**a 2 1 = (r 2 – r 1) / **|**r 2 – r 1**|

**r 1**= Location of Q 1
**r 2** = Location of Q 2
r 2 1= | r 2 – r 1 | (Magnitude)
**ε 0** = Permittivity of free space = 8.854 x 10 – 12, F /

m & = 9 x 10 9

The force on Q 1 due to Q 2 is **F 1 = – F 2 **8

Problem 2-4: If Q 1 and Q 2 are two point charges and are located at r 1 and r 2 respectively in free space. Find out The force on Q 2 due to Q 1? The force on Q 1 due to Q 2?

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Q 1 = First charge in free space.
Q 2 = Second charge in free space.
**r 1**= Location of Q 1
**r 2** = Location of Q 2

Distance vector along the line joining two charges

**r 2 1 = (r 2 – r 1)
**

The magnitude of **r 2 1 **is obtained as; r 2 1** = **|**r 2 – r 1**|

Now the direction of **r 2 1 **is obtained as; **r 2 1 **= **r 2 1 / **r 2 1 = **a** 2 1

The force on Q 2 due to Q 1 in free space is written as in the form of; **F 2 = { } a** 2 1

The force on Q 1 due to Q 2 in free space is written as in the form of; **F 1 = { } a** 1 2

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Here;

**r 1 2 = (r 1 – r 2) ** => ** – (r 2 – r 1) **=>** – r 2 1
**

**r 2 1 = (r 2 – r 1) **=> ** – (r 1 – r 2)
**

r 1 2 = r 2 1

=>r 1 2** = **| **r 1 – r 2 **| => r 2 1** = **|**r 2 – r 1**|

**a r 1 2 ** = ( **r** 1 2** / **r 1 2** ) **=> **( r 1 – r 2 ) / **| **r 1 – r 2 **|

**a r 2 1** = ( **r** 2 1 ** / **r 2 1 ** ) **=> (– **r** 1 2 ** / **r 1 2 ** ) **=>– **( r 1 – r 2 ) / **| **r 1 – r 2 **| => **a r 1 2 ** = – **a r 2 1 **=> **F 1 **= – **F 2 **

Problem 2-6: A point charge Q 1 = 2 µ C is at (2, 3, 6) and another charge Q 2 = 5 µ C is at (0, 0, 0) in free space. Find out the force on Q 1 due to Q 2?

Q 1 = 2 x 10 – 6 C

Q 2 = 5 x 10 – 6 C

**r 1 **= 2** a x + **3** a y + **6** a Z
**

**r 2 **= 0** a x + **0** a y + **0** a Z
**

Hence,

**r = (r 1 – r 2) **= > 2** a x + **3** a y + **6** a Z
**

The magnitude of **r **is obtained as;

r **= **| **r **| = > (x 2 + y 2 + z 2) ½

= > {(2) 2 + (3) 2 + (6) 2}½ = > (49) ½ = > r = 711

The unit vector **r **is obtained as;

**a 1 2 = r / **r= >{ 2 a x + 3 a y + 6 a z} / 7
In this way; **F 1 2 = ( ) a** 1 2

= >F 1 2 = { } { }

= > F 1 2 = {(9 x 10 9 x 10 x 10 – 6 ) / 343} { 2 a x + 3 a y + 6 a z}

= >F 1 2 = 0.26 x 10 – 3 { 2 a x + 3 a y + 6 a z}

= >** F 1 2** = (0.52 a x + 0.78 a y + 1.56 a z) N.

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Problem 2-7: Three equal point charges of 2 µ C are in free space at (0, 0, 0), (2, 0, 0) & (0, 2, 0) respectively. Find out net force on Q 4 = 5 µ C at (2, 2, 0)?

Q 1 = Q 2 = Q 3 = Q = 2 x 10 – 6 C

Q 4 = 5 x 10 – 6 C

**r 1**= (0, 0, 0) = >

**r 2** = (2, 0, 0) = > 2 a x

r 3 = (0, 2, 0) = > 2 a y

r 4 = (2, 2, 0) = > 2 a x + 2 a y

r 4 1 = (r 4 – r 1) = > 2 a x + 2 a y

r 4 2 = (r 4 – r 2) = > 2 a x + 2 a y – 2 a x = > 2 a y

r 4 3 = (r 4 – r 3) = > 2 a x + 2 a y – 2 a y = > 2 a x 13

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The magnitude of **r 4 1 **is obtained as;

r 4 1=|**r 4 – r 1**| = > (x 2 + y 2 ) ½

= > {(2) 2 + (2) 2} ½ = > {(8)} ½ = >r 4 1** = 2.828
**

r 4 2** = **|**r 4 – r 2**| = > {(0) 2 + (2) 2} ½

= > {(4)} ½ = > r 4 2** = 2
**

r 4 3** = **|**r 4 – r 2**| = > {(2) 2 + (0) 2} ½

= > {(4)} ½ = > r 4 3** = 2**

The net force on Q 4 due to Q 1, Q 2 and Q 3 is ; F = F 4 1 + F 4 2 + F 4 3

F 4 1 ={ } a 4 1 => a 4 1 =

F 4 2 = { } a 4 2 => a 4 2 = => a 4 2 = a y

F 4 3 = { } a 4 3 => a 4 3 = => a 4 3 = a x

F = { }

=> { } = > 9 x 10 9 x 2 x 10 – 6 x 5 x 10 – 6 { }

= > 3.98x10 – 3 {7.65 a x + 7.65 a y} = > 30.44 (a x + a y) m N

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Problem 2-8: Two point charges of Q 1 = 2 µC and Q 2 = 5 µC are located at (– 3, 7, – 4) and (2, 4, – 1) respectively. Determine the force on Q 2 due to Q 1 and the

force on Q 1 due to Q 2 ?

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