8/19/2019 PHY103_Lec_5

1/13

8/19/2019 PHY103_Lec_5

2/13

Summary of Lecture # 4:

2

• Gauss’s Law

() =1

40�r 2   ̂ 

Φ =  ⋅

 

⋅ =  0 

• Scalar Potential : if × = 0 everywhere, = −V • Vector Potential : if ⋅ = 0 everywhere, =  ×  

•Coulomb’s Law:

• Electric Flux

8/19/2019 PHY103_Lec_5

3/13

Gauss’s Law from Coulomb’s Law:

3

() = 140�  ̂r 2  ′  

Coulomb’s law gives the electric field due a volume charge ′  

If Coulomb’s and Gauss’s law have same information

content, can we derived Gauss’s law from Coulomb’s law?

Take the divergence of both sides of the equation

⋅ () = 140� ⋅   ̂r 2  ′  

We have: ⋅    ̂r 2 = 4 ( ) = 4 (− ) 

⋅ = 140�4  − ′   ′  = ()0  

Therefore,

⋅ =  0 The divergence of electric field is equal to the charge density

divided by 0 

8/19/2019 PHY103_Lec_5

4/13

Applications of Gauss’s Law:

4

Q: (Griffiths: Ex 2.10): What is the flux through the shaded face of the cube due to the

charge

 at the corner

Ans:

⋅

=1

24 0 

8/19/2019 PHY103_Lec_5

5/13

Curl of the Electric Field:

5

() =1

40 r 2   ̂ Electric field due to a single point charge  is:Let’s take the simplest electric field:

We need to find the curl of it

× = 40 ×1

r 2   ̂  

Take the area integral

� × ⋅

=40� ×

1

r 2   ̂ ⋅  

� × ⋅ = 40� ×1

r 2   ̂ ⋅ = 401

r 2   ̂ ⋅  Use Stokes’s theorem

� × ⋅

=

401

r 2 = 40 ×

1

 

= 0  × =  

8/19/2019 PHY103_Lec_5

6/13

Curl of the Electric Field (Digression):

6

× =  Curl of an electric field is zero. We have shown this for the simplest field, which is the field of

a point charge. But it can be shown to be true for any electric field, as long as the field is static.

What if the field is dynamic, that is, what if the field changes as a function of time?

×

=

 −

  Faraday’s Law in differential form.

Integrate over a surface

� × ⋅

= � − ⋅  Apply Stokes’ theorem

� ⋅

= −  � ⋅  

E= −Φ

 EMF

Magnetic flux Faraday’s Law in integral form form.

8/19/2019 PHY103_Lec_5

7/13

Maxwell’s Equations (Digression 2):

7

× = −   � ⋅ = −  � ⋅  

⋅ =  0  ⋅ =enc0  

⋅ = 0 

× = 0 − 00  

Gauss’s Law

Faraday’s Law

Amperes’s Law with

Maxwell’s correction

 No name; Magnetic

Monopole does not exist

When the fields do not vary as a function of time, it is called Electrostatics / Magnetostatics.

(before mid-sem)

When fields do vary as a function of time, then the two fields have to be studied together

as electromagnetism (Or electrodynamics) and one consequence of a changing electric

and magnetic field is the electromagnetic radiation. (after mid-sem)

When the energy of the field is quantized (photons) then it is called quantum electrodynamics.

(Not for this course). Applications: Quantum computers, Quantum cryptography, Quantumteleportation

8/19/2019 PHY103_Lec_5

8/13

Electric Potential:

(1) 

 ∫  ⋅d

b

a  is independent of path.

(2) ∮ ⋅ = 0 for any closed loop. This is because of Stokes’ theorem� × ⋅ =  ⋅    

• This is because Curl of a gradient is zero × V =  (3)  is the gradient of a scalar function: = −V 

If the curl of a vector field   is zero, that is, if × = 0 everywhere, then:

8

Recall:

(1)  ∫  ⋅ dba   is independent of path.(2)

∮ ⋅ = 0 for any closed loop.

This is because of Stokes’ theorem

� × ⋅ =  ⋅

 

 

(3)  is the gradient of a scalar function: 

The curl of Electric field   is zero, that is, if × = 0 everywhere. Therefore:

V  is called the electric potential. It is a scalar quantity,

the gradient of which is equal to the electric field

= −V 

8/19/2019 PHY103_Lec_5

9/13

Electric Potential:

9

Use the fundamental Theorem for Gradient:

How to write electric potential in terms of the electric field?

−V =  Take the line integral of the above equation over a path

�V ⋅

= −� ⋅

 

V  − V() = −� ⋅

 

� ⋅ =  − ()

 

 

 

• Absolute potential cannot be defined.

• Only potential differences can be defined.

Since  × = 0 everywhere, = −V 

8/19/2019 PHY103_Lec_5

10/13

Electric Potential:

10

(4) V  − V() = −∫  ⋅ . Absolute potential cannot be defined. In electrostatics,usually one takes the reference point to infinity and set the potential at infinity to zero,that is, take V  = V ∞ = 0. Also if V  = V(),

V

 =

 − � ⋅

 

(1) Electric potential is different from electric potential energy. Unit of electric potential

is Newton-meter per Coulomb (N⋅mC  ) or Volt.

(2) The potential obeys the principle of superposition, that is, the potential to a set ofcharges is equal to the sum of the potentials due to individual charges: V = V1 + V2 +⋯ (3) Whatever can be calculated using electric fields can also be calculated using electric

 potential. However, electric field is a vector quantity whereas electric potential is a

scalar quantity. So, doing calculations with potential is relatively easier. Although

electric potential is a scalar quantity, one can get all the information about the electricfield (a vector quantity) because × = 0, that is

Ex =

 Ey ;

Ez =

 Ey ;

Ex =

 Ez ;

8/19/2019 PHY103_Lec_5

11/13

Electric Potential due to a point charge at origin:

11

(

) =

1

4012

 

 

Electric field () at  due to a single point charge  at origin:

V  = − � ⋅

 

Electric potential V   at   due to a single point charge  at origin:

The line element is: = 1 r + 11  + 1sin11   

V  = − �1

40 12

1 = −   40 �1

12

1 = 401

 

V  =

401

 

= − �() ⋅

 

8/19/2019 PHY103_Lec_5

12/13

Electric Potential due to a localized charge distribution:

V  = 401

r 

 

V  = 140r i

=1 

V  =

401

 

Potential due to a point charge  at origin

Potential due to a point charge  at  

Potential due to a collection of point charges

V() = 140�r  For a line charge =  ′  For a surface charge

=

  ′

  

For a volume charge =  ′  

Potential due to a a continuous charge distribution is

12

8/19/2019 PHY103_Lec_5

13/13

Ease of calculating the Electric Field

• If the above two is not applicable, one has to go back to the Coulomb’s law and

then calculate the electric field.

• The easiest is to calculate the electric field using Gauss’s law. But this is

 possible only when there is some kind of symmetry in the problem.

• The next best thing is if the electric potential is known then one can calculate

the electric field by just taking the gradient of the potential = −V .Sometimes, it is even beneficial to calculate the electric potential first as an

intermediate step and then calculate the field from there.

13