CLASS 12 PHYSICS, MODULE – 01, ELECTROSTATIC,

CHAPTER – 04, Electrostatic Potential

* Electrostatic Potential :

The electrostatic potential at a point in the electric field is defined as the workdone in bringing a unit positive charge from infinity to that point against the electrostatic force.

Units :

1 ) SI unit is Volt or J / C

2 ) CGS unit is stat – volt

Dimension : ML2 T-3 A-1

* Electrostatic Potential Difference :

Electrostatic Potential Difference between two points in the electric field is defined as the work done to move a unit positive charge from one point to another point in the electric field against the electrostatic force due to the field.

Units :

1 ) Its SI unit is Volt or J / C

2 ) Its CGS unit is stat – volt

Dimension : ML2 T-3 A-1

* Electric Field as a gradient of electric potential :

Let potential at point P = V volt

Potential at point Q = ( V + dV ) volt

q0 = Test charge

E = Electric field at point P due to charge +q placed at point O.

Force experienced by test charge ( F ) = – q0

Small displacement ( PQ ) = d

workdone due to displacement PQ ( dW ) = F . d

( here bold letters indicate vector representation )

⇒ dW = – q0 E . d

⇒ dW / q0 = – E . dr

⇒ { ( V + dV ) – V } = – Edr cos0          ( E and dr are parallel to each other )

⇒ dV / dr = – E

Here, dV / dr is the potential gradient and it is equal to Electric field and negative sign indicates that E is directed always in the direction of decrease of electric potential.

dV / dr = – E 

* Potential at a point due to a point charge :

Let a point charge is brought at point P from infinity and the point P is at a distance of r from point O at whic +q charge is kept. The electric potential at point P is given by

V = – E . dr

[ here E is the electric field produced due to charge +q and is equal to Kq  / r2

and  K = 1 / ( 4πε0 ) = 9 * 109 Nm2C-2

and ε0 is the absolute permitivity of free space ]

⇒ V = – ∫ Edr cos0

⇒ V = – ∫ Edr 

⇒ V = – ∫ ( Kq  / r2 ) dr 

⇒ V =   Kq / r + c               ( c is integral constant )

at r = ∞ , V = 0

0 = 0 + C

⇒ c = 0

therefore V = Kq / r

* Electric Potential due to Electric Dipole :

Electric potential at different positions of the electric dipole will be different.

1 ) Electric Potential at a point on axial position of the electric dipole :

Let there is an electric dipole of length 2l and dipole moment p = ( 2l * q). Let us consider that there is a point P at a distance of ‘x ‘  on the axial position of the electric dipole.

The electric potential due to charge +q at point P is given by

V+q = Kq / ( x + l )

The electric potential due to charge -q at point P is given by

V-q = – Kq / ( x – l )

where  K = 1 / ( 4πε0 ) = 9 * 109 Nm2C-2

and ε0 is the absolute permitivity of free space

Now, electric potential at point P due to the electric dipole is given by

V = V+q + V-q

⇒ V =  Kq / ( x + l ) – Kq / ( x – l )

⇒ V = Kq ( -2l ) / ( x2 – l2 )

⇒ V = K ( q * 2l ) / ( l2 – x2 )

⇒ V = kp /  ( l2 – x2 )

2 ) Electric potential at point on the perpendicular bisector of the electric dipole :

Let there is an electric dipole of length 2l and dipole moment p = ( 2l * q). Let us consider that there is a point P at a distance of ‘x ‘  on the perpendicular bisector of the electric dipole.

The electric potential due to charge +q at point P is given by

V+q = Kq / √ ( x2 + l2 )

The electric potential due to charge -q at point P is given by

V-q = – Kq / √ ( x2 + l2 )

where  K = 1 / ( 4πε0 ) = 9 * 109 Nm2C-2

and ε0 is the absolute permitivity of free space

Now, electric potential at point P due to the electric dipole is given by

V = V+q + V-q

⇒ V = Kq / √ ( x2 + l2 ) – Kq / √ ( x2 + l2 ) = 0

Hence, the electric potential at a point P on the perpendicular bisector of the electric dipole is zero.

* Electric potential due to a charged spherical conductor ( hollow or solid ) :

Let a charge +q is given to sphere of radius R . Let us assume tat the whole charge is concentrated at the center O. Let us consider that there is a point P outside the sphere at a distance of r.

Electric potential at point P is given by

V = Kq / r

where  K = 1 / ( 4πε0 ) = 9 * 109 Nm2C-2

and ε0 is the absolute permittivity of free space.

Conclusion :

i ) If r = R , that is electric potential at a point on the surface of the charged sphere is given by

V = Kq / R = constant

ii ) If r < R, that is the point lies inside the charged sphere.

As we know that, there is no electric lines of force inside the conductor, therefore

electric field = 0

⇒ dV / dr = 0

⇒ dV = 0

⇒ V = constant

Hence, electric potential at any point or on the surface of the charged sphere is remains constant.

* Electric potential due to an uniformly charged circular ring :

Let

a = radius of circular ring

λ = linear charge density = charge per unit length

x = distance of the point P at which the potential is calculated from the center of the ring.

dq = elementary charge = λ dl

dl = elementary length

dv = elementary electrostatic potential = K λ dl / { √ ( a2 + x2 ) }

where  K = 1 / ( 4πε0 ) = 9 * 109 Nm2C-2 and ε0 is the absolute permittivity of free space.

Now, the electrostatic potential due to the circular ring is given by

v = ∫ dv 

v = { K / √ ( a2 + x2 ) } ∫ λ dl

v = qK / { √ ( a2 + x2 ) }

Electric potential energy :

Electric potential energy of a system of charges is defined as the work done required to assemble this system of charges by bringing them from infinite distance.

# Electric potential energy of a system of two point charges :

Let q1 and q2 are the charges placed at infinite distance from each other. q1 is brought at a point A . But no work is done because no electrostatic force due to any other charges opposes the motion of q1 .

Therefore, work done ( w1 ) = 0

Now, q is brought at another point B from infinity. Its motion is opposed by the electric field produced by the charge q1 and thus work has done and given by

W2 = q2 * V1

⇒ W2 = q2 * ( Kq1 / AB )

⇒ W2 = Kq1q2 / r12

Therefore, the electrostatic potential energy due to a system of two point charges is given by

U = W1 + W2

U = Kq1q2 / r12

# Electrostatic Potential Energy due to a system of three point charges :

Let q1 , q2 and q3 are the charges placed at infinite distance from each other. q1 is brought at a point A . But no work is done because no electrostatic force due to any other charges opposes the motion of q1 .

Therefore, work done ( w1 ) = 0

Now, q is brought at another point B from infinity. Its motion is opposed by the electric field produced by the charge q1 and thus work has done and given by

W2 = q2 * V1

⇒ W2 = q2 * ( Kq1 / AB )

⇒ W2 = Kq1q2 / r12

If the third charge q3 is brought to a point C from infinity. Its motion is opposed by the electric fields produced by charges q1 and q2. Hence, the work done in bringing the third charge form infinity to the point C is given by

W3 = q3 ( VB + VA ) = Kq2q3 / r23 + Kq1q3 / r13

Therefore, the electrostatic potential energy due to a system of three point charges is given by

U = W1 + W2 + W3

U = 0 + Kq1q2 / r12  + Kq2q3 / r23 + Kq1q3 / r13

U = ( Kq1q2 / r12  ) + ( Kq2q3 / r23 ) + ( Kq1q3 / r13

Conclusion : The electrostatic potential energy due to a system of n point charges is given by

U = ( Kq1q2 / r12  ) + ( Kq2q3 / r23 ) + ( Kq1q3 / r13 ) + ………………………..+ Kqn-1qn / r1( n – 1 ) n

# Kinetic Energy of a charged body :

Let us assume that a charged body having charge q  is moving in a electric field of electric potential V, with velocity v. Here, work done by the charge is equivalent to its kinetic energy.

Therefore, qV = mv2 /2                ( m is the mass of the charged body )

v = √ ( 2Vq / m )

For electron, q = e

v = √ ( 2Ve / m )

Here, e /m is called as specific charge and its value is 1.77 * 1011 C / kg.

* Equipotential Surface :

Any two dimensional surface which has same electric potential at each and every point on the surface, is termed as the Equipotential surface.

It is defined as the locus of all the points having same electric potential.

# Properties of Equipotential Surface :

1 ) Electric field is perpendicular to equipotential surface.

2 ) Its shape depends on the charge the charge distribution.

3 ) No two equipotential surfaces intersect each other.

4 ) Work done on this surface is zero.

5 ) Electric lines of force are perpendicular to this surface.

6 ) Electrostatic potential at each and every point is equal.

 


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