Transcript Electric Potential

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
I, IV, III, II

F
II, I, IV, III
III, II, IV, I
I
II
I, IV, II, III


F
III, IV, I, II
F
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
force
and
displacement vectors:

F

F

r

r
II
I
WII   Fr
WI  0

F

F
 
W  F  r  Fr cos

r

r
III
WIII  F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
 mgyi  mgy f

yf
yi
Then
U g  mgy


r

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 energy of the system
Uf
U  U f  U i  W
The work done by the electrostatic
force is path independent.
 Work done by a electric force or “field”

Ui
 
 
W  F  r  qE  r

Uf
Work done by an Applied force
K  K f  K i  Wapp  W
Wapp  W
Ui
U  U f  U i  Wapp
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
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.
f
October 3, 2007
i
E
Electric Potential

The electric potential energy



Start
Then
So
 
dW  F ds

dW  q0 E  ds
f 

W  q0  E  ds
i
U  U f  U i  W  q0 
f
i

The electric potential
V  V f  Vi 
Uf
q

U
V
q
 
E  ds

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.
U i U

q
q
f 

U
V 
   E  ds
i
q0
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 U  W and V  W / q at any point in an 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 for
for
+
q
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 always 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.
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. WI, 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.
B.
C.
D.
E.
WI = WII
WIII is not equal to zero
WII equals to zero
WIII = WIV
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

So
i
R
1
q
40 r 2
q

1
q
0 V  
dr

40 R r 2
40
V (r ) 
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
V  Vi 
i 1
n
qi

40 i 1 ri
1
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,
V   dV 
1
dq
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

then,

So
dq  dx
1 dq
1
dx
dV 

40 r
40 ( x 2  d 2 )1/ 2

dx

2
2 1/ 2 L



ln
x

(
x

d
) 0
2
2 1/ 2
4

(
x

d
)
4

0
0
0
L
V   dV  



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
B 


VB  VA   E  ds  0
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 q0 moves through a displacement ds
from on equipotential surface to the adjacent surface.
 The work done by the electric field on the test charge is W = dU = -q0 dV.
 
 The work done by the electric field may also be written as W  q0 E  ds
dV
 Then, we have
 q0 dV  q0 E (cos  )ds
E cos   
ds

So, the component of E in any direction is the negative
of the rate at which the electric potential changes with
distance in that direction.


If we know V(x, y, z),
Ex  
V
x
Ey  
V
y
Ez  
Es  
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
40
(
q1q2 q1q3 q2 q3


)
r12
r13
r23
October 3, 2007
Summary


Electric Potential Energy: a point charge moves from i to
U  U f  U i  W
f in an electric field, the change in electric potential
energy is
Electric Potential Difference between two points i and f in
U
U
U

V

V

V



an electric field:
q
q
q
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
1 q
to corresponding equipotential surfaces.
V (r ) 
40 r
f 

Finding V from E:
U
V 
   E  ds
i
q0
n
Potential due to point charges:
1 n qi
V  Vi 

Potential due to a collection of point charges:
40 i 1 ri
i 1
Potential due to a continuous charge distribution:
1
dq
Potential of a charged conductor is constant everywhere V   dV 
40  r
inside the conductor and equal to its value to its value at
the surface.
V
V
V
V
E 
E 
E 
E 
1 q1q2
Calculatiing E from V:
y
z
x
s
U  q2V 
40 r
Electric potential energy of system of point charges:
f
f








s
x
y
i
z
October 3, 2007
i