Coulombs Law Pdf study matterial, Lecture notes for Electronics engineering
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Coulombs Law Pdf study matterial, Lecture notes for Electronics engineering

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Coulombs Law and related problems
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Single phase controlled rectifiers

Coulombs Law:

Subject: Electro Magnetic Field Theory.

Course In charge: Engr. M.Asghar Daudani.

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2

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  

12r

12F

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 0ε 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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