Electricity & Magnetism – Lecture 6

Electric Potential
Conservative forces
A conservative force “gives back” work that has been done against it

Work done on the charge (by an

external agent, or by the field) will
result in changes in the potential
energy of the charge.

Gravitational and electrostatic forces are conservative

Electric Potential Energy
Electrostatic force
Gravitational force

∆   -  = -W
Note: Electric energy is one type of energy.

work W done to accelerate a positive charge from rest is

positive and results from a loss in PE, or a negative ΔU. There
must be a minus sign in front of ΔU to make W positive.
Potential energy in a constant field 

 The potential energy difference between A

and B equals the negative of the work
d q0 done by the field as the charge q0 is moved
dl A from A to B
B

∆   -  = -

Potential energy difference between A and B

∆   -  = - ⃗.   −   . 
 

But E = constant, and E.dl = -1 E dl, then:

∆ = -   .  =   =    =   d ∆    d
  
Electric potential difference
The potential energy ∆U depends on the charge being moved. In order to remove this
dependence, the concept of electric potential ∆V is introduced

Electric Potential = Potential Energy per Unit Charge

∆VAB = ∆UAB / q0

∆VAB = VB – VA
Electric potential difference between the points A and B

∆  
∆ = =- ( )    .  = - .  ∆! = - 
."#
  
Cases in Which the Electric Field  is not Aligned with "#

∆! = - 
."#


A
θ

Since F = q E is conservative, the field E is conservative. Then, the electrical potential difference does not
depend on the integration path.

One possibility is to integrate along the straight line AB.

This is convenient in this case because the field E is constant, and the angle θ between E and dL is constant.


∆! = − cos ' 
dl
Electric potential
• The electric potential difference ∆ in volts between two points
is the work in Joules needed to move 1 C of charge between
those points

The 1.5 V battery does 1.5 J of

work for every 1 C of charge
flowing round the circuit
Reference Point of Electric Potential Energy
∆   -  = -W
The reference point can be anywhere. For convenience, we usually set charged particles
to be infinitely separated from one another to be zero potential energy
The potential energy U of the system at any point f is

  −f
where W∞ is the work done by the electric field on a charged particle as that particle moves in from
infinity to point f.
f joule/coulomb=volt (V)
−

•Both the electric potential energy U and the electric potential V are scalars.
•The electric potential energy U and the electric potential V are not the same. The electric potential energy is
associated with a test charge, while electric potential is the property of the electric field and does not depend
on the test charge.
Electric Potential Due to a Point Charge
  
E=
()* + ,

∆! = - 
."#
+q B dl A
. =Ecos /   −  rB
A test charge is moved through a distance dl rA
to the left, actually it moved in the direction
of decreasing r.
− .
+
.   −     .  −  = - 
.  = - +
 .
 + 0+   
 −  = -  ( - )
()* + + , ()* + +
  
If rA tends to ∞ then VA=0 V=
()* +

 2
For group of point charges V= ∑
()* +2

These are not vector quantity

 0
For continuous charges V= 
()* +
Electric potential energy
1 q1
V=
4πε 0 r
q2
r12
1q1q2 .
U = q2V =
4πε 0 r
q1
r23

r13
If the system consists of more than two charged
particles, calculate U for each pair of charges and
sum the terms algebraically
q3
1 q1q2 q1q3 q2 q3
U = U12 + U13 + U 23 = ( + + )
4πε 0 r12 r13 r23
Equipotential Surface
Lines joining the equal potential are Equipotential lines.

In three dimensions, the lines form equipotential

surfaces. Any surface over which the potential is
constant is called an equipotential surface. In other
words, the potential difference between any two points
on an equipotential surface is zero.

On a contour map, the curves mark constant elevation; the steepest slope is
perpendicular to the curves. The closer together the curves, the steeper the
slope.
Equipotential Surface
Properties Equipotential Surface (1) No work is done in moving a test charge
from one point to another on an equipotential
 Equipotential surfaces are always perpendicular to surface. (2) The electric field is always
electric field lines. perpendicular to the element dl of the
Because no work is done to move the charged particle, It equipotential surface. (3) Equipotential
must always be perpendicular to field lines. Hence
equipotential surface is always perpendicular to field lines. surfaces indicates regions of strong or weak
 No work is done by the electric field on a charged electric fields. (4) Equipotential surfaces
particle while moving the particle along an cannot intersect.
equipotential surface.

Constant E Point Charge

Electric dipole
Electric potential at any point along the axial line of an electric dipole

Electric Potential due to an electric dipole at axial point.

Consider an electric dipole AB, having charges - q and + q

at points A and B respectively.

The separation between the charges is 2l.

Electric dipole moment, p = q . 2l, directed from - q to + q.
Consider a point P on the axis of dipole at a distance r from
mid-point O of dipole.
The distance of point P from charge + q is BP = r -l
The distance of point P from charge - q is AP = r + l
Let V1 and V2 be the potentials at P due to charges + q and
- q respectively.
  3
V=
()* + ,
Electric Potential Due to Electric Quadrupole
  4
V= Q= quadrupole moment
()* + 5
It however does have a non-vanishing quadrupole moment
Find electric potential due to an electric dipole at point p
Problem
Find the Potential at the center of the square.
N N
1 qn
V (r) =  Vn (r) = 
n =1 4 πε o n =1 rn

q1 = +12 nC q2 = -24 nC
+ -
d = 1.3 m

+ +
q4 =+31 nC q3 =+17 nC
Problem
Figure shows three point charges held in fixed positions by forces that are not shown. What
is the electric potential energy U of this system of charges? Assume that d=12 cm and that
Problem
Potential energy U of a system of point charges
q2
y We define U as the work required to assemble the
r12 system of charges one by one, bringing each charge
r23 from infinity to its final position
q1 Using the above definition we will prove that for
a system of three point charges U is given by:
r13
q3
x
O q1q2 qq qq
U= + 2 3 + 1 3
4πε o r12 4πε o r23 4πε o r13

Note: Each pair of charges is counted only once

Problem
Find the mutual potential energy
N N
1 qn
V (r) =  Vn (r) = 
n =1 4 πε o n =1 rn

q1 = +12 nC q2 = -24 nC
+ -
d = 1.3 m

+ +
q4 =+31 nC q3 =+17 nC