# CLASS 12 PHYSICS, MODULE – 01, ELECTROSTATIC,

### CHAPTER – 02, ELECTRIC FIELD

Electric field : It is defined as the space around a charge where another charge can experience force due to former charge. Electric field

From coulomb’s law of electrostatic, we have

F = K q1q2 / r2

Where K is proportionality constant which is given by

K = 1 / ( 4πε0 ) = 9 * 109 Nm2C-2

where ε0 is the absolute permitivity of free space.

If q1 and q2 are non zero and F = 0 then

r = ∞

that is the mathematically, the electric field is extended up to ∞ but practically its extension is limited.

Electric Field Intensity :

From Coulomb’s law we have

F = K q1q2 / r2

Where K is proportionality constant which is given by

K = 1 / ( 4πε0 ) = 9 * 109 Nm2C-2

where ε0 is the absolute permitivity of free space.

If q1 = q and q2 = + 1 unit then

E = q / r2

The electric field intensity at a point in the electric field is defined as the eletrostatic force experienced by unit positive charge placed at point . It is vector quantity. Electric Field Intensity

Relation Between The Electrostatic Force And Electric Field Intensity :

From Coulomb’s law of electrostatic we have

F = K q1q2 / r2

Where K is proportionality constant which is given by

K = 1 / ( 4πε0 ) = 9 * 109 Nm2C-2

where ε0 is the absolute permittivity of free space.

The electric field intensity is given by

E = K q2 / r2

From above two equations we can write

F = q1 E

The required relation between the force and electric field intensity is given by

F = qE

Here, F and E are in the same direction for positive charge while F and E are in opposite direction for negative charge.

Important points regarding the Electric Field :

1 ) Charge particle in an electric field experiences force if the particle is at rest or moving.

2 ) In presence of medium or dielectric around the charge or charge distribution, electric field intensity at any point around it will be 1 / K times less than field in air, where K is dielectric constant.

3 ) If identical charges are placed at each of the vertices of regular polygon, then net electric field at center of polygon is zero.

4 ) Test charge used to measure the magnitude and direction of the electric field intensity must be positive.

5 ) The electric field is conservative in nature.

Electric Dipole :

An electric dipole is said to be formed if equal and opposite charges are separated by a small distance. Electric Dipole

Electric Dipole Moment :

The product of length of the dipole and the either of the two charges is called as the electric dipole moment. Electric Dipole Moment

p = q * 2l

where q is the magnitude of the charge of the electric dipole and 2l is the length of the dipole.

Unit :

1 ) Its SI unit is C- m

2 ) Its CGS unit is stat coulomb – cm

Electric Lines Of Force :

It is a path along which a free positive charge moves and tangent drawn at any point on the path gives the direction of the electric field. Electric lines of force

Properties Of Electric Lines Of Force :

1 ) These are the imaginary lines originating from the positive charge and end on negative charge.

2 ) The direction of the electric field at a point on the electric lines of force is given by the tangent drawn at that point.

3 ) Two electric lines of force can not intersect each other because at the point of intersection two tangents can be drawn and this will indicate the two different directions of the electric field at the same point simultaneously. This is an impossible situation. So, the two electric lines of force can never intersect each other.

4 ) The electric lines of force having tendency to contract along their length. This explains the attraction between two opposite charges.

5 ) The electric lines of force having tendency to separate in a direction perpendicular to their length. This shows the repulsion between two same charges.

6 ) Closer the distance between the lines of force higher is the electric field intensity and vice versa.

7 ) No electric lines of force exist inside the conductor.

8 ) This electric lines of force are imaginary lines but they, represent the electric field is real.

9 ) The electric lines of force are perpendicular to the equipotential surface.

10 ) These lines of force are always perpendicular to the surface of the charged body.

11 ) These lines of are discontinuous curves.

12 ) These lines don not form any closed loop.

13 ) These lines are either straight or curved paths.

14 ) The magnitude of the field is represented by the density of field lines. Denser the lines of the force indicates the strong electric field and vice versa.

15 ) The electric lines of force originates from positive charge and terminates to a negative charge.

Effect of Electric Field on point charge :

If a point charge q is kept in a electric field, it experiences the electrostatic force F due to electric field intensity E. Due to this force, it accelerates.

As we know that

F = qE

ma = q E

a = ( q / m ) E

This is the acceleration through which the point charge of mass ‘ m ‘ accelerates.

( q / m ) is the specific charge of the particle.

Expression for the Electric Field Intensity due to an Electric Dipole :

Let us discuss the electric intensity due to presence of electric dipole at three different position.

1 ) Expression for the electric field intensity due to an electric dipole moment on the axial position  / tanA position / end on position : Electric field intensity due to an electric dipole at axial position

Let a unit positive charge is placed at a point P at a distance distance of ‘r’ from an electric dipole of length ‘2l’ and dipole moment ‘ p ‘

p = 2l * q

where q is the either of the charges

Let E+q = electric intensity due to +q = kq / ( r – l )2

E-q = electric intensity due to -q = kq / ( r +  l )2

Where

K = 1 / ( 4πε0 ) = 9 * 109 Nm2C-2

where ε0 is the absolute permitivity of free space.

Resultant electric intensity at point P is given by

E = E+q – E-q

E = {  kq / ( r –  l )2 } – {  kq / ( r +  l )2 }

E = kq { ( r + l )2 – ( r – l )2 } / { ( r + l )2 ( r – l )2 }

E = Kq { 4rl / ( r2 – l2 )2 }

E = { k / ( r2 – l2 )2 } * ( 2l * q ) * 2r

E = 2kpr / ( r2 – l2 )2

This is the expression for the electric intensity due to an electric dipole on axial position.

Conclusion :

1 ) It is directed along the axis of the electric dipole.

2 ) It is proportional to the electric dipole moment of the electric dipole.

3 ) For a small dipole r >>>l, then in the above expression ‘ l ‘ may be neglected.

E = 2kpr / ( r2 – l2 )2

E ≅ 2kpr / ( r2 )2

E = 2kp / r 3

2 ) Expression for the electric field intensity due to an electric dipole moment on the perpendicular bisector position  / tanB position / broadside on position :  Let a unit positive charge is placed at a point P at a a perpendicular distance from an electric dipole of length ‘2l’ and dipole moment ‘ p ‘

p = 2l * q

where q is the either of the charges

sine components of the electric field intensity at point P are acting in opposite direction so, they  cancel each other while its cosine components are in the same direction, so they will give the resultant electric intensity.

Let E+q = electric intensity due to +q = E-q = electric intensity due to -q = kq / √( r2 +  l2 )2

Where

K = 1 / ( 4πε0 ) = 9 * 109 Nm2C-2

where ε0 is the absolute permitivity of free space.

cosθ = l / √( r2 + l2

Resultant electric intensity at point P is given by

E = E+q cosθ  + E+q cosθ

E = 2E+q cosθ

E = 2 * { kq / √( r2 +  l2 )2 } * { l / √( r2 + l2 ) }

After calculating, we will get

E = kp / ( r2 + l23/2

Conclusion :

1 ) The electric field intensity acts parallel to the axis of the dipole ( from negative to positive )

2 ) For very short dipole r >>>>l, then in the above expression, l can be neglected,

therefore, E = kp / r3

3 ) The electric field intensity is proportional to the electric dipole moment.

Expression Of The Torque In The Electric Field : Expression for the Torque experienced by the electric dipole

Torque experienced by an electric dipole placed in an electric dipole is given by

τ = Force * perpendicular distance

τ = q E * x

τ= q E * 2 l sinθ

τ = ( q * 2 l ) * E sinθ

τ = p E sinθ

τ = p × E

where θ is the angle of inclination of the electric dipole, q is the amount of the charges on the ends of the electric dipole and 2 l is the length of the dipole and E is the electric field intensity.

Hence, the torque experienced by an electric dipole placed in an electric field is the cross product of the electric dipole moment and the electric field intensity.

If the electric field is uniform then the dipole will experience only torque. If the electric field is non uniform, then it will experience both the force as well as the torque.

Conclusion :

1 ) τ = p E sinθ, if θ = 0 degree i.e the dipole is placed parallel to the electric field.

Work done in rotating an electric dipole :

Let us consider that the electric field is uniform and the torque experienced by the dipole is given by

τ = p E sinθ

work done in rotating the electric dipole from θ1 to θ2 is given by

W = ∫ τ .dθ

W = ∫ p E sinθ dθ

W = pE θ2θ1  sinθdθ

W = pE [ -cosθ ] θ2θ1

W = pE [ – { cosθ2 – cosθ1 } ]

W = pE ( cosθ1 – cosθ2

Conclusion :

1 ) If the dipole is initially oriented parallel to the electric field and it is rotated through the angle θ, then the work done is given by

W = p E ( 1 – cosθ )

2 ) If the dipole is perpendicular to the electric field and it is rotated through angle θ, then the work done is given by

W = p E ( cos 90 – cosθ )

W = – p E cosθ

3 ) If the dipole makes the angle θ with the electric field and it is rotated through the angle 90 degree, then the work done is given by

W = p E cosθ

W = p . E

Objective type questions :

1 ) Is electric field is vector or scalar quantity ?

Answer : Electric field is vector quantity.

2 ) What is the dimension of electric field intensity if energy ( U ), electric charge ( Q ), velocity ( v ) and time ( T ) are considered as the fundamental quantities ?

where F is the electrostatic force, q is the electric charge and E is the electric field.

[ E ] = [ F ] / [ q ] = [ work / displacement ] / [ Charge ]

[ E ] = [ energy / ( velocity * time ) ] / [ charge ]

[ E ] = { U / ( VT ) } / Q

[ E ] = U V-1 T-1 Q-1

3 ) What is the value of electric field intensity inside a conductor ?

Answer : Zero as there is no electric lines of force inside a conductor.

4 ) Which physical quantity is represented by the electric lines of force ?

Answer : Electric field is represented by the electric lines of force.

5 ) What will be the direction of electrostatic force experienced by a unit positive charge placed at a point inside a electric field created by a positive charge ?

Ans : The electrostatic force and the electric field has the same direction.

6 )   If identical charges are placed at each of the vertices of regular polygon, then what will be the electric field at center of polygon ?

7 ) Are the electric lines of force are continuous curve ?

Answer : No, the electric lines of force are discontinuous curve.

8 ) The electric lines of force are always __________ to the surface of the charged body. Fill in the blank.