8/19/2019 PHY103_Lec_6

1/12

Semester II, 2015-16

Department of Physics, IIT Kanpur

PHY103A: Lecture # 6(Text Book: Intro to Electrodynamics by Griffiths, 3rd  Ed.) 

Anand Kumar Jha

1

8/19/2019 PHY103_Lec_6

2/12

Summary of Lecture # 5:

2

•Curl of the electric field : × = 0 

• Electric Potential: = −V 

• Electric Potential due to a localized charge distribution

• Gauss’s Law from Coulomb’s Law: ⋅ =   

V  − V() = −� ⋅

 

V  = − � ⋅

 

V  = 401r 

 

Point charge Charge distribution

V() = 140 �r  

8/19/2019 PHY103_Lec_6

3/12

3

Charge distribution in terms of electric potential:

Gauss’s Law ⋅ =   

But × = 0 ⇒ = −V 

⋅ =  ⋅ −V  Therefore,

2V = −  0  Poisson’s Equation

In the region of space where there is no charge, =0

2V = 0  Laplace’s Equation

= − ⋅ V  = −2V =  

0 

8/19/2019 PHY103_Lec_6

4/12

4

Summary: 

 

V    = −V 

V  = − � ⋅

 

Electrostatics

8/19/2019 PHY103_Lec_6

5/12

Work and Energy in Electrostatics 

There is a charge  in an electrostatic field . Howmuch work needs to be done in order to move the

charge from point  to ?

= � ⋅

 = −� ⋅

 

•   = − is the force one has to exert in order tocounteract the electrostatic force = . 

= Q V  − V() 

• Work is independent of the path.

Take V()=V(∞)=0  and V  = V    = QV   

• Work done to move a unit charge from point  to  is the potential difference between points  and  

• Work done to create a system of unit charge (to bring a unit

charge from ∞ to  is the electric potential.If Q = 1, = V   •

Thus, electric potential is the potential energy per unit charge5

8/19/2019 PHY103_Lec_6

6/12

The work required to create a system of a point

charge  is:

Work required to bring in the charge 1 from ∞ to   = QV  . 

1 = 1V0 = 1 × 0 = 0 Work required to bring in the charge 2 from ∞ to  

2=

 2V

1 

Work required to bring in the charge 3 from ∞ to  3 = 3V2 

Total work required to bring in the first three charges:

= 1 +2 +3 =   1r 

+ 

r +

 r 

 

Total work required to bring in the first four charges:

=  1

r  +

 r  +

 r  +

r  +

 r  +

 r    6

Work required to assemble  point charges: 

= 2  1

r 

 

= 3   1r 

+ r   

8/19/2019 PHY103_Lec_6

7/12

Total work required to bring in the first four charges:

=   1r  +

 r  +

 r  +

r  +

 r  +

 r   

Total work required to bring in  point charges, with charge 1, 2, 3 ⋯, respectively. :

= 140

r 

>

=1 

=1

2×

1

40

r 

=1≠

=1 

= 12× 140   r 

=1≠

=1 

W =1

2

=1

() • This is the total worked required to assemble  point charges• The potential

(

) is the potential at

 due to all charges,

except the charge at . 7

Work required to assemble  point charges: 

8/19/2019 PHY103_Lec_6

8/12

 

W =1

2

=1() 

• This is the total worked required to assemble  point charges• The potential () is the potential at  due to all the

other charges, except the charge at

.

The Work required to assemble (or the energy of ) a continuous charge Distribution:

What would be the required work if it is continuous distribution of charge ?

W =1

2

=1  

Well…

Is this correct?→   1

2�  

 

→  1

2�      Not really !The potential ()  inside the integral is the potential at point . However, the

 potential  inside the summation in the potential at  due to all the chargesexcept the charge at . Because of this difference in the definition of the

 potentials, the integral formula turns out to be different. 8

8/19/2019 PHY103_Lec_6

9/12

 

W =1

2�  

  =02 � ( ⋅ )

 

W = − 02 � ⋅ + 02 � ⋅  

W = − 02 � ⋅

+

02  ⋅

 

Using the product rule

⋅  =  ⋅ + ⋅ () Using the divergence theorem

∫   ⋅ =

 ∮   ⋅ 

 

W =02 � 2

+

02  ⋅

  Using - =  

W =02 �

2

  

When the volume we are integrating over is very

large, the contribution due to the surface integral is

negligibly small.

Using = 0( ⋅ ) 

9

The Work required to assemble (or the energy of ) a continuous charge Distribution:

8/19/2019 PHY103_Lec_6

10/12

 The Energy of a Continuous Charge Distribution: 

W =1

2�  

  W =02 � 2

  

So, what is the energy of a point charge using the above formula?

Electric field of a point charge is

The energy of a point charge is therefore,

W = 02 � 1402

r 2 2 

2sin   = 280 � 12 

   

= ∞  ??W =

1

2

=1

 →   12�    due to incorrect conversion

of the sum into an integral

This is the total energy of a charge distribution

including the self energy of assembling the charge

distribution. Assembling a point charge requires

infinite energy. This is why this expression gives

infinity for the energy of a point charge.

This is the total work done to

assemble a set of point

charges. This does not

include the self energy of

assembling a point charge.

10

= 140r 2   ̂ 

8/19/2019 PHY103_Lec_6

11/12

 The Electrostatic Energy (Summary): 

The work required to create a system of a point charge  is: = QV   

= 12

=1() 

() is the potential at  due to allcharges, except the charge at

.

W =1

2�  

 

=02 �

2

  

Total work required to put together  point charges is: 

The Energy of a Continuous Charge Distribution:

 Note # 1: The total work required to assemble a continuous charge distribution

11

= the total energy of a continuous charge distribution

 Note # 2: The self energy of assembling a point charge is infinite. Therefore, the

total energy of a point charge is infinite.

8/19/2019 PHY103_Lec_6

12/12

 Where is the electrostatic energy stored ?

W =1

2�  

 

=02 � 2

  

The Energy of a Continuous Charge Distribution:

• The first expression has a volume integral of   over the localized chargedistributions whereas the second one has the volume integral of 2over allspace.

• Just as both the integrals are mathematically correct, both the interpretations

are also correct. The electrostatic energy can be interpreted as stored locally

within the charge distribution or globally over all space.

• Again, at this point, as regarding fields, we know how to calculate different

 physical quantities but we don’t really know what exactly the field is. 12

So, where is the electrostatic energy stored? Within the charge

distribution or over all space?