Slide 1

Electromagnetic FieldCoulombs Law Coulomb's law states that the force between two point charges and is: 1. Directly proportional to the product the charges and . 2. Inversely proportional to the square of the distance (R )between them.

R

;

Coulombs law in vector form 1In vector form

By coulombs law

. 3

. 2Eqn. 2 and Eqn. 3 in Eqn. 1

- Unit vector in the direction of force ;

;

3

From the diagram

Force due to N number of chargesIf there are there charges Q1, Q2, Q3 , with position vector r1,r2,r3,..rN respectively as shown in the bellow figure.The total force on Q due to all the charge is equal to the vector sum of the force on Q due to each charges.By coulombs law the force acting on due to

Force on Q due to the charges Q1 ,Q2 , and Q3

Similarly if there are N no. of charges electric field due to al the charges :

Therefore the electric field (E) at a point P due to the charge :Electric Field Intensity

The force acting on the unit positive charge:

The Electric field (E) at a point is defined as , the force acting on the unit positive charge at that point

By coulombs law , the force (F) on the charge Q due to the charge :

r

6Electric Field due to N number of charge If there are there charges Q1 ,Q2 ,Q3 with position vector r1,r2,r3 respectively as shown in the bellow figureThe electric field (E) at a point P due to the charge :

By superposition principle, The total electric field (E) at point P, due to all the charges is equal to , the vector sum of the electric field at the point P due to individual charges.

The electric filed due to all the three charges: Similarly The electric filed due to N no. of charges:

7 Point charges 1 mC and - 2 mC are located at (3, 2, - 1 )m and (1, 1,4)m, respectively in free space. Calculate the force on a 10-nC charge located at (0,3,1)m.

= 1mc= -2 mc= 10 nc(3,2,-1)(-1,-1,4)(0,3,1)

The force on Q due to N number of charges

For free space

Consider a line from a point A to B with uniform charge density as shown in figureThe line is lies in the z axis Consider a small length dz of the line.Let the charge present in the small length (dz) of line is

Electric field at a point P due to small length (dz) of line,

- the vector connecting the small length dz and the point P

- the unit vector in the direction

Electric field due to the line ,from A to B, at point P ,

....3Derivation for Electric Field due to charges distributed uniformly on an infinite line 1Eqn. 2 in Eqn. 1

2

P

(0,0,z)AB

Changing the integration with respect to angle

From the figureFrom the figure

.A.BEqn.A / Eqn.BD.wt.r to z .4.5.6.7.8Eqn.4,Eqn.5,Eqn.6,Eqn.7 and Eqn.8 in Eqn.1

P

(0,0,Z)(0,0,Z1)

11

Substitute the above values in Eqn.9For infinite line

9A uniform line charge of 16 nC/m is located along the line defined by y = 2, z = 5. If Find E at P(1, 2, 3)

5(x,-2,5)P(1,2,3)

-2

-Line charge density in C/M

-Perpendicular distance between line charge and point in Meter

-Unit vector in Meter

Given the potential field and a point P(-4,3,6) find the following values at a point P:1.Potential(V),2.Electric field(E), 3.Flux density(D).

1.Potential ( ):2.Electric field( ):

We know that

3.Flux density(D).

Given a field v/m find potential difference between A(-7,2,1) and B(4,1.2)

We know thatThe line integral from point A to B can be split into three paths as bellow

Path.1: (4,1,2) to (-7,1,2) x alone variesPath.2: (-7,1,2) to (-7,2,2)- y alone variesPath.3: (-7,2,2) to (-7,2,1)- z alone varies