physics121 lecture05

Physics 121: Electricity &
Magnetism – Lecture 5
Electric Potential
Dale E. Gary
Wenda Cao
NJIT Physics Department


Work Done by a Constant
Force
1.

A.
B.
C.
D.
E.

The right figure shows four situations in which a force is
applied to an object. In all four cases, the force has the
same magnitude, and the displacement of the object is to
the right and of the same magnitude. Rank the situations
in order of the work done by the force on the object, from
most positive to most negative.

F

F
I, IV, III, II
II, I, IV, III
I

II
III, II, IV, I


F
I, IV, II, III
F
III, IV, I, II
III

IV
October 3, 2007




Work Done by a Constant
Force 

The work W done a system
by an agent exerting a
constant force on the
system is the product of
the magnitude F of the
force, the magnitude Δr of
the displacement of the
point of application of the
force, and cosθ, where θ is
the angle between the
  displacement

force and
W F r Fr cos 
vectors:

F


F


r


r
II

I

WII  Fr

WI 0


F


F
III



r

WIII Fr


r
IV

WIV Fr cos 
October 3, 2007


Potential Energy, Work and
Conservative Force


Start

 
Wg F r  mgˆj [( y f  yi ) ˆj ]


mg


r

mgyi  mgy f

yf

yi



Then
U g mgy





The work done by a conservative
force on a particle moving between
any two points is independent of
the path taken by the particle.



The work done by a conservative
force on a particle moving through
any closed path is zero.

So
Wg U i  U f  U
U U f  U i  Wg

October 3, 2007


Electric Potential Energy



The potential
system

energy

of

the

Uf

U U f  U i  W
Ui

The
work
done
by
the
electrostatic
force
is
path
independent.
 
 
W Fby
ra electric

qE r force or
 Work done
“field”


K K f  K i Wapp  W


Work done by an Applied force
Wapp  W
U U f  U i Wapp

Uf
Ui

October 3, 2007


Work: positive or negative?
2.

In the right figure, we move the proton from point
i to point f in a uniform electric field directed as
shown. Which statement of the following is true?

A.

Electric field does positive work on the proton; And
Electric potential energy of the proton increases.
Electric field does negative work on the proton; And

f
Electric potential energy of the proton decreases.
Our force does positive work on the proton; And
Electric potential energy of the proton increases.
Electric field does negative work on the proton; And
Electric potential energy of the proton decreases.
It changes in a way that cannot be determined.

B.
C.
D.
E.

October 3, 2007

i

E


Electric Potential


The electric potential energy








Start dW F ds
 
Then dW q0 E ds
So
f 


W q0  E ds
i

 
U U f  U i  W  q0  E ds
f

i



The electric

U
potential
V
q

U f U i U
V V f  Vi 



q
q
q

V 

f 

U
  E ds
i
q0



Potential difference depends only
on the source charge distribution
(Consider points i and f without
the presence of the test charge;



The difference in potential energy
exists only if a test charge is
moved between the points.

October 3, 2007

Electric Potential


Just as with potential energy, only differences in electric
potential are meaningful.








Relative reference: choose arbitrary zero reference level for ΔU or ΔV.
Absolute reference: start with all charge infinitely far away and set Ui
= 0, then we have
at any point in an
U  W
V and
 W / q
electric field, where W is the work done by the electric field on a
charged particle as that particle moves in from infinity to point f.

SI Unit of electric potential: Volt (V)
1 volt = 1 joule per coulomb
1 J = 1 VC and 1 J = 1 N m
Electric field:
1 N/C = (1 N/C)(1 VC/J)(1 J/Nm) = 1 V/m
Electric energy:
1 eV = e(1 V)

= (1.60×10-19 C)(1 J/C) = 1.60×10-19 J

October 3, 2007


Potential Difference
in a Uniform Electric Field








Electric field lines always point in
the direction of decreasing electric
potential.
A system consisting of a positive
charge and an electric field loses
electric potential energy when the
charge moves in the direction of
the field (downhill).
A system consisting of a negative
charge and an electric field gains
electric potential energy when the
charge moves in the direction of
the field (uphill).
Potential difference does not
depend on the path connecting

them

downhill
uphill forfor

+qq

f 
f
f

V V f  Vi   E ds   ( E cos 0 )ds   Eds
i

i

i

f

V V f  Vi  E  ds  Ed
i

U q0 V  q0 Ed
c 
c

Vc  Vi  E ds  ( E cos 90 )ds 0
i
i

f 
f
f

V f  Vi   E ds   ( E cos 45 ) ds  E cos 45  ds
c

c

V f  Vi  E cos 45

c

d
 Ed
sin 45

October 3, 2007


Equipotential Surface


The name equipotential surface is given to
any surface consisting of a continuous
distribution of points having the same electric
potential.



Equipotential
surfaces
are
perpendicular to electric field lines.



No work is done by the electric field on a
charged particle while moving the particle
along an equipotential surface.

always

Analogy to Gravity


The equipotential surface is like the “height”
lines on a topographic map.



Following such a line means that you remain
at the same height, neither going up nor
going down—again, no work is done.
October 3, 2007


Work: positive or negative?
3.

The right figure shows a family of equipotential surfaces
associated with the electric field due to some distribution
of charges. V1=100 V, V2=80 V, V3=60 V, V4=40 V. W I,
WII, WIII and WIV are the works done by the electric field on
a charged particle q as the particle moves from one end to
the other. Which statement of the following is not true?

A.

WI = WII

B.

WIII is not equal to zero

C.

WII equals to zero

D.

WIII = WIV

E.

WIV is positive
October 3, 2007




Potential Due to a Point
Charge

Start with (set Vf=0 at  and Vi=V at R)

f 
f


V V f  Vi   E ds   ( E cos 0 )ds   Edr
i



We have

E




Then



R

1 q
4 0 r 2

q
0  V 
4 0

So

V (r ) 


i

1
q
dr

R r 2
4 0


E

1 q
4 0 r 2



1 q
1



r 
4 0 R
 R

1 q
4 0 r

A positively charged particle produces a
positive electric potential.
A negatively charged particle produces a
negative electric potential
October 3, 2007


Potential due to
a group of point charges


Use superposition
n
 
r 
 n
V  E ds   Ei ds  Vi
r





i 1



i 1

For point charges
n

1
V  Vi 
4 0
i 1

n

qi

i 1 ri

The sum is an algebraic sum, not a vector sum.
 E may be zero where V does not equal to zero.
 V may be zero where E does not equal to zero.


q

q

q

-q

October 3, 2007


Electric Field and Electric
Potential
4. Which of the following figures have V=0
and E=0 at red point?
q

q

q

-q

A

B

q

q

q

-q

q

q

q

q

C

D

q

-q
E

October 3, 2007


Potential due to a Continuous
Charge Distribution


Find an expression for dq:






dq = λdl for a line distribution
dq = σdA for a surface distribution
dq = ρdV for a volume distribution

Represent field contributions at P due to
point charges dq located in the distribution.

dV 


1 dq
4 0 r

Integrate the contributions over the whole
distribution, varying the displacement as
needed,

1
dq
V dV 
4 0  r

October 3, 2007


Example: Potential Due to
a Charged Rod


A rod of length L located along the x axis has a uniform linear
charge density λ. Find the electric potential at a point P located on
the y axis a distance d from the origin.



Start with



L
then, V  dV  
 

dq dx
1 dq
1
dx
dV 

4 0 r
4 0 ( x 2  d 2 )1/ 2

0




So

dx


ln x  ( x 2  d 2 )1/ 2
2
2 1/ 2
4 0 ( x  d )
4 0

 



L
0



ln  L  ( L2  d 2 )1/ 2   ln d 
4 0
 L  ( L2  d 2 )1/ 2 

V
ln 

4 0 
d

October 3, 2007

Potential Due to
a Charged Isolated Conductor





According to Gauss’ law, the charge resides on
the conductor’s outer surface.
Furthermore, the electric field just outside the
conductor is perpendicular to the surface and
field inside is zero.
Since

 
VB  VA   E ds 0
B

A





Every point on the surface of a charged
conductor in equilibrium is at the same electric
potential.
Furthermore, the electric potential is constant

everywhere inside the conductor and equal to
its value to its value at the surface.
October 3, 2007


Calculating the Field from the
Potential



Suppose that a positive test charge q 0 moves through a
displacement ds from on equipotential surface to the adjacent
surface.
 
The work done by the electric field on the test charge W
is W
=

q0 EdUd=s -q0
dV.
dV

q
dV

q
E
(cos

)

ds
E
cos



0
0
The work done by the
electric
field may also be written as
ds
Then, we have



E 
So, the component of E in any direction is the negatives






of the rate at which the electric potential changes with
distance in that direction.


V
E



x
If we know V(x, y, z),
x

E y 

V
y

E z 

V
s

V
z

October 3, 2007


Electric Potential Energy
of a System of Point Charges
U U f  U i  W

 
 
W F r qE r

Wapp  W

U U f  U i Wapp



Start with (set Ui=0 at  and Uf=U at r)
1 q1
V
4 0 r



We have

U q2V 


q2

q1

1 q1q2
4 0 r

If the system consists of more than two
charged particles, calculate U for each pair of
charges and sum the terms algebraically.

U U12  U13  U 23 

1 q1q2 q1q3 q2 q3
(


)
4 0 r12
r13
r23
October 3, 2007


Summary














Electric Potential Energy: a point charge moves from i
to f in an electric field, the change in electric potential
energy is

Electric Potential Difference between two points i and
f in an electric field:
Equipotential surface: the points on it all have the
same electric potential. No work is done while moving
charge on it. The electric field is always directed
perpendicularly to corresponding equipotential
surfaces.
f 

U
Finding V from E:
V 
  E ds
i
q0
Potential due to point charges:
Potential due to a collection of point charges:
Potential due to a continuous charge distribution:
Potential of a charged conductor is constant
everywhere inside the conductor and equal to its
value to its value at the surface.
V
V
V
V
E 
E z 
E x 
Es 
Calculatiing E from V:

y
z
x
s
Electric potential energy of system of point charges:
y

U U f  U i  W
U
U
U
V V f  Vi  f  i 
q
q
q

1 q
4 0 r

V (r ) 
n

V  Vi 
i 1

1
4 0

n

qi

i 1

i

r

1
dq
V dV 
4 0  r
U q2V 

October 3, 2007

1 q1q2
4 0 r