AP Physics: Electricity & Magnetism
What is potential energy?
The ability to do work
The energy possessed by
an object due to its
position in a force field.
Work is an energy transfer
A constant force applied over a
distance in the same direction.
W  Fd

W

sf
si
F  ds  U  U i  U f
The work done by a conservative force
equals the decrease in the potential energy
of the particle.

W

W

B
A
Fc  ds  U
B
A
q 0E  ds  U
U  q 0  E  ds
B
A

This is called a line or path integral which is
independent of the path taken from point A to B.

B
U
   A E  ds  V
q0
The potential at
any point in the
field:
This value is called
Electric Potential
and is independent
of the value of q
U
V
q0
Be careful, electric potential is not
potential energy.
Units?
J/C
The work per unit charge that an external agent
must perform to move a test charge from A to B
without a change in kinetic energy
VP    E  ds
P

The electric field at an arbitrary point equals
the work required per unit charge to bring a
positive test charge from infinity to that point
B
U
   A E  ds  V
q0

J  N 
  m V


C  C 
 Joule  Newton meter
 
 Volt 



Coulomb  Coulomb 

The energy that an electron or proton gains or loses
by moving through a potential difference of 1V
1.60210 J 1eV
19

What is the potential difference between two points if
the displacement is parallel to the field lines?
V  VB  VA    E  ds
B
A
   E cos 0ds    Eds  E  ds 
B
B
B
A
A
A
V  Ed
VB  VC
Cross-section of
equipotential
surface
A
E
B
C
The name given to any
surface consisting of a
continuous distribution of
points having the same
electric potential.
No work is done in moving a test charge between any
two points on an equipotential surface.
 Red
lines are electric field lines
 Blue lines are equipotential surfaces
 Equipotential surfaces must be perpendicular
to field lines.
A negative charge gains electric potential
energy when it moves in the direction of the
electric field. Explain this.
A proton loses potential energy when moving
in the direction of the electric field, but
picks up an equal amount of kinetic energy.
Work done by a field is positive when energy
is given to an object from the field (object is
lowered in a gravitational field or moved in
the direction of E filed lines). This occurs
when Uf <Ui so, W=-ΔU
V  Ed
d  0.0006m
E  3 106 V /m
V  Ed
V  3 106 V /m0.0006m

V  1.8kV

U
V 
q
V  120V
v0  0
v?
mp  1.67 1027 kg
me  9.111031 kg
e  1.60219 C

1
qV  U  KE  mv2
2

2qV
v
m
vp 1.5 10 m/s

5
ve  6.5 106 m/s
qV  U  KE
V  115V
KE
q
V
KE  7.371017 J
q?

q  6.4 1019 C

VA  9V
VB  5V
q  N ae  (6.021023 )(1.6021019 C)
q  96440.4C
N a  6.02 1023 electrons
W ?

V  VA  VB
V  9V  (5V)
V  14V
W  U  qV
 W  qV  (96440.4C)(14V)
W 1.35MJ
What is the electric field due to a point charge (q)?
Q in
 E  dA  
0
Field is radially outward (or inward), and
the Gaussian surface is a sphere chosen to
match symmetry of field.
Q in
EA 
0
Q in
q
kq
E

 2
2
A0 4r 0 r
kq
E 2
r
V  VB  VA    r E  ds
rB
A
The following simplification can be made because the
potential difference is not dependant upon the path
taken through
 the field, only the endpoints.
V    r Edr
rB
A
kq
V    r 2 dr
A r

rB
1
V  kq  r 2 dr
A r
rB
 1rB
V  kq 
 r rA
rB
 1
V  kq 
 r rA
If we want to know the potential at one particular location
near a point charge we chose the first point to be infinity.
1rB
V  kq 
r 
1 1 
V  kq  
r 

kq
V
r
kq
V
r
Potential at a point due to many
point charges.
qi
V  k
i ri
This is a scalar value so it is much
easier to calculate than E field.
The potential energy of a
pair of charged particles
separated by a distance r12.
this is also the work done to
move q2 from infinity to
point P where q1 is located.
kq1q 2
U  q 2V1 
r12
V1 is the electric potential at point
P due to charge q1.

Notice that if the charges are like then positive work
is done to bring the charges together (they have more
energy together because they will repel). Energy
delivered to the system requires positive work.
Imagine q1 is at a fixed position. We want to bring q2
and q3 from infinity to a position near q1.
q1q2 q1q3 q2q3 
U  k



r13
r23 
 r12
Each term represents the the work required to bring each charge
to a location near the other. It does not matter which order we
bring them in from infinity.
r12
q1
q2
r13
r23
q3
V    E  ds
B
A
How can we use this to solve for E field?
dV  Eds

If the E field only has one component…


dV  E x dx
dV
Ex  
dx
dV
Ex  
dx
Remember that the potential is zero for displacements
perpendicular to the field
If 
your potential is a function of all three spatial
coordinates (x, y, z)
dV
Ey  
dy
dV
Ez  
dz
V  4x y  y  2yz
2
2
To find the electric field, we must take the
partial derivative of this potential…
V 
Ex 
 4x2y  y2  2yz
x x
Partial derivative: Treat other variables as constant
while differentiating with respect to one variable.
 2
 2  4y2x  8xy
E x  4x y  4y x
x
x
V  4x y  y  2yz
2
2
V 
2
2
Ey 
 4x y  y  2yz
y y
E y  4x2  2y  2z
V 
Ez 
 2yz
z z
E z  2y

E x,y,z  8xyˆi  4x2  2y  2zˆj  2ykˆ
In vector notation this is often written as…
  ˆ   ˆ   ˆ
E  V   i   j   k
x  y  z 
 is called the gradient operator
If the potential is constant in some region, what
is the electric field?
If the electric field is zero in some region, what
is the electric potential?
If the potential is constant in some region, what
is the electric field?
If the electric field is zero in some region, what
is the electric potential?
v  constant, E  0
E  0, v  constant
The difference in electric potential (voltage)
measured when moving from point A to point B is
equal to the work which would have to be done,
against the electric field to
per unit charge,
move the charge from A to B.
The electric potential at point P due to a
continuous charge distribution can be calculated
by dividing the charged body into segments of
charge dq and summing the potential
contributions over all segments.
kq
V
r
dq
Vk
r
dq
a
P
x
dq
Vk
r
r a x
2
2
All segments of charge are equidistant from point P
V
k
a x
2

2

 dq
V
kQ
a x
2
2
dq
a
P
The field is zero here,
therefore the potential
must be constant:
kQ
V
a
this potential is the work necessary to bring a
test charge from infinity to the location in the
center of the ring.
This oddly shaped
conductor with an
excess positive charge
is in equilibrium
A
B

 VB  VA    E  ds  0
B
A
Eds  0
Along this surface path
E is always
perpendicular to the
displacement ds.
The surface of any charged conductor in equilibrium is
an equipotential surface. Furthermore, since the
electric field is zero inside the conductor, we conclude
that the potential is constant everywhere inside the
conductor and equal to its value at the surface.
A
B
The electric field is large near points having small
convex radii of curvature and reaches very high
values at sharp points.
B
Because the potential on the
surface of the cavity is an
equipotential surface.
A
A cavity in a conductor with no charge in it…The
electric field inside the cavity must be zero,
regardless of the charge distribution on the
outside surface of the conductor.
 Sensitive
circuits and people can be
protected if placed in a cavity inside of a
conductor.
Air can become a conductor as
a result of the ionization or air
molecules in regions of high
electric fields. At STP this
happens around 3x106 V/m
+2Q
.
E
. . .
.
A
D
B
C
-Q
2a
-Q
Three charges are arranged in an equilateral
triangle, as shown. At which of these points
(a,b,c bisect sides, d is equidistant from other
points) is the electric potential smallest?
+2Q
.
E
. . .
.
A
D
B
C
-Q
2a
-Q
C) A small positive test charge will move
towards an area of low potential. Ask yourself
“Where would an small positive charge end up
if released near these charges?”
.P
10V
40V
70V
The diagram shows a set of equipotential
surfaces. At point P, what is the direction of
the electric field?
A) left
B) right
C) up the slide
D) down the slide
E) either left or right, which one cannot be determined
.P
10V
40V
70V
A) A small positive charge placed at P would
move to a location of low potential (left). The
force that moves it is caused by an electric
field which will be in the same direction or
opposite it. Equipotential surfaces are always
perpendicular to electric field lines.
a
Q
Q
What is the electric potential at a point
halfway between the two charges?
A) kQ/a
B) 2kQ/a
D) Electric potential
is a scalar
C) zero
D) 4kQ/a
E) 8kQ/a
V
kQ kQ 2kQ 4kQ



a
a
a
a
2
2
2
A solid conducting sphere carries a
charge +Q. Which of the following is
true of the electric field E and the
electric potential V inside the sphere?
A) E=0
and
V=0
B) E=0
and
V≠0
C) E≠0
and
V=0
D) E≠0
and
B) The field in a conductor is
always zero. E is the
derivative of V, so to be zero,
V must have been a constant.
V≠0
E) It cannot be determined without
knowing the radius of the sphere.
dV
E 
dr
A negatively charged rod is brought near a metal
object on an insulating stand, as shown. When charges
stop moving, the left side of the object has an excess
of positive charge, and the right side of the object,
where the radius of curvature is less, has an excess of
negative charge. Which of the following best describes
the electric potential on the metal object?
+
-------
+
+
+
-
A) It is greatest on the + side
B) It is greatest on the – side
C) It is greatest at the center
D) It is the same everywhere on the object
E) It cannot be determined from
the
information given
+
-------
+
+
+
-
A) It is greatest on the + side
B) It is greatest on the – side
C) It is greatest at the center
D) It is the same everywhere on the object
E) It cannot be determined from
the
information given
D) It is the same everywhere on the object.
The surface of a conductor is an
equipotential surface. The E field is 0 inside
the conductor and V is constant.
.
L
.
Q
L
L
Q
L
.
Q
L
L
.
Q
Four identical charged particles, each with
charge Q, are fixed in place in the shape of an
equilateral pyramid with sides of length L, as
shown above.
What is the potential energy of this
arrangement of charges?
.
L
In general:
q1q 2
U
40r
.
Q
Q
L
L
L
.
Q
L
L
.
Q
Each pair of charges is separated by a distance L.

Each pair contributes:
Q2
U
40L
There are 6 pairings:
6Q 2
U
40L

 What
is the amount of work required to
assemble 4 identical point charges of
magnitude Q at the corners of a square of
side s?
1 1 1 1
1
1 
U  kQ     


s s s s
2s
2s 
2
Q
s
Q

2s
s


Q
Q 
4
2 
U  kQ  

s
2s 
2
kQ2 
2 
U
4 

s 
2 
kQ 2
U  5.41
s
A wire that has a uniform linear charge density
λ is bent into the shape shown below.
Find the electric potential at point O.
2R
R
.
O
2R
A positron is given an initial velocity of
6x106m/s. It travels into a uniform electric
field, directed to the left. As the positron
enters the field, its electric potential is zero.
What is the electric potential at the point
where the positron has a speed of 1x106m/s?
+
vi
E