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Electric Potential Energy
h
Think of gravitational potential energy.
When the block is moved vertically up
against gravity, the gravitational force does
negative work (you do positive work), and the
potential energy (U) increases.
When the block falls vertically down, the
gravitational force does positive work, and
the potential energy (U) decreases.
a
b
PHYS 153 08W 1PHYS 153 08W 1PHYS 153 08W 1
baab UUW −=mghUUU ab =−=∆In this case,
When is positive (block falling), baW → ba UU > , and is negative.U∆
The potential energy decreases.
When the block is lifted up, the gravitational work is negative, and
is positive.
U∆
Gravitational force is a conservative force.
The work done by a conservative force has 4 properties.
1. It is reversible.2. It is independent of the path between the start and end points.
3. When the start and end points are the same, total work is zero.
4. The work can be expressed as a difference in potential energy.
The electric field force is conservative, and the ideas above apply to the work
done in moving a charge in an electric field.
For a positive charge moving
in the direction of E, U
decreases, and moving
opposite to E, U increases.
PHYS 153 08W 2PHYS 153 08W 2PHYS 153 08W 2
Fig. 23.3
opposite to E, U increases.
The opposite is true for a
negative charge.
For a conservative force, the
change in K.E. is minus the
change in P.E.
)( abab UUKK −−=− bbaa UKUK +=+or
Can easily show that because the electrostatic force is conservative, the work
done by this force is independent of path.
When a force acts on a particle that moves from point a to point b, the
work done is given by a line integral,
→
F
∫∫ ==→→
→
b
a
b
aba dlFldFW φcos.
→
ld is the incremental displacement along the particle’s path, and φis the angle between
→
F and →
ld at any point.
PHYS 153 08W 3PHYS 153 08W 3PHYS 153 08W 3
is the angle between F and ld at any point.
Fig. 23.6
In the figure, a and b do not lie on the same
radial line.
∫ ∫==→
b
a
b
aba EdrqdlFW 0cosφ
since drdl =φcos
So, the work done during a small
displacement →
ld depends only on the change dr in the radial distance r,
independent of the path.
(a) Electric potential energy in a uniform field.
+
+
+
+
+
-
-
-
-
-
•q0 →
F
dWork done over some distance d is
EdqFdW 0==
Work done by the field is positive, so
the potential energy of the particle
PHYS 153 08W 4PHYS 153 08W 4PHYS 153 08W 4
+
+
-
-
x0
the potential energy of the particle
decreases.
In going from a to b, the decrease in
a b00ExqP.E. is
(b) Electrical P.E. of two point charges.
Consider a positive test charge q0 a distance r from a positive charge q.
Here the force is not constant with
distance as in the uniform field.
If the force moves the particle from
a to b along a radial line
drr
qqrdFW
b
a
b
aba ∫∫ ==
→→
→ 2
0
04
1.
πε
)11
(4
0
rr
qqUW −=∆−=
πε
PHYS 153 08W 5PHYS 153 08W 5PHYS 153 08W 5
)(4 0 ba rr
UW −=∆−=πε
The work, and hence change in P.E., depends
only on the end points. Here the work is positive
and the P.E. change is negative.
Recall, the work done does not depend on the path
between the end points.
Suppose we have infinity as our reference point, where U=0 and bring q0from infinity to point a.
Fig. 23.5
Then
ar
qqU 0
04
1
πε= )0( =∞U
Thus, the potential energy U of a test charge q0 at any distance r from charge
q is
r
qqU 0
04
1
πε= (electric potential energy of two point charges
q and q0.)
(c) Electric potential energy of several point charges.
PHYS 153 08W 6PHYS 153 08W 6PHYS 153 08W 6
(c) Electric potential energy of several point charges.
Fig. 23.8
Recall the resultant electric field is the
superposition of all individual fields. The
work done to move q0 to point a is the sum
of the work done against each field. So the
P.E. of charge q0 at point a is the algebraic
sum of the potentials of each pair of charges
involving q0.
∑=+++=i i
ia
r
r
q
r
q
r
qqU
0
0
3
3
2
2
1
1
0
0
4...)(
4 πεπε
If q0 is moved to another point b, then Ub is given by the same expression
with the distances ri measured to point b. The work done in moving q0 from
a to b is ba UU −
The work done in bringing together all the charges in Fig. 23.8 (not just q0)
is given by
∑qq1
PHYS 153 08W 7PHYS 153 08W 7PHYS 153 08W 7
∑<
=ji ij
ji
r
qqU
04
1
πε(i<j so that we include each pair only once)
Electric Potential
Potential is the potential energy per unit charge. The symbol is V, and the
unit is the volt.
0q
UV = V, or J/C.
Refer back to Fig. 23.3 on slide 2, and use the equations on slide 1.
The work done by the field in moving the positive charge from a to b,
baab UUW −=
The electric field does positive work, and the potential energy decreases.
The potential is given by abbabaab VVVq
U
q
U
q
W=−=−=
000
The work done by the field in moving the charge from a to b is positive,
And so Va > Vb.
PHYS 153 08W 8PHYS 153 08W 8PHYS 153 08W 8
(a) Potential of a single point charge.
Recall the expression for the electric potential energy of two point
charges, q and q0.
Hence r
q
q
UV
00 4
1
πε==
What is the reference point here?
(b) Potential due to a collection of point charges.
Recall the expression for the electric potential energy for a point charge q0and a collection of charges qi.
Hence ∑==i i
i
r
q
q
UV
00 4
1
πεPotential at point a a distance ri from qi.
Potential here is calculated with respect to the same point in (a). This point is infinity, where
V=0.
(c) Potential due to a continuous distribution of charge.
PHYS 153 08W 9PHYS 153 08W 9PHYS 153 08W 9
(c) Potential due to a continuous distribution of charge.
∫=r
dqV
04
1
πεr is the distance from dq to the point where V is being
evaluated, again with respect to inifinity.
•q0Recall, potential is the work per
unit charge.
Suppose a positive charge q0 is
moved from a to b.
a
b→→
→ ∫= ldEqWb
aba .0
∫∫ ==−∴→→ b
a
b
aba dlEldEVV φcos.
In this case is negative, and hence Vb > Va .baW →
We see from this that the units of E can be V/m as well as N/C.
PHYS 153 08W 10PHYS 153 08W 10
Problem 22.48
A solid conducting sphere with raduis R carries positive total charge Q
The sphere is surrounded by an insulating shell with inner radius R and
outer radius 2R. The insulating shell has a uniform charge density .
(a) Find the value of so that the net charge of the entire system is zero.
(b) If has the value found in part (a), find the magnitude and direction of
electric field in each of the regions 0 < r < R, R < r < 2R, r > 2R. Plot a
graph of the radial component of E as a function of r.
(c) Do the results agree with the general rule that the electric field is
discontinuous only at locations where there is a thin sheet of charge?
ρρ
ρ
PHYS 153 08W 11PHYS 153 08W 11
Examples of calculating Electric Potential
Problem 23.32
A total electric charge of 3.50 nC is distributed uniformly over the surface of a
metal sphere with a radius of 24.0 cm. If the potential is zero at a point at
infinity, find the value of the potential at the following distances from the centre
of the sphere:
(a) 48.0 cm, (b) 24.0 cm, (c) 12.0 cm.
Recall that the electric field due to a uniformly
charged sphere is given by
PHYS 153 08W 12
Fig. 23.17
charged sphere is given by
2
04
1
r
qE
πε= (same as for a point charge at the
centre).
The potential due to a point charge is
r
qV
04
1
πε=
Problem 23.33
A uniformly charge thin ring has radius 15.0 cm and a total charge of 24.0 nC.
An electron is placed on the ring’s axis a distance of 30.0 cm from the centre
of the ring and is constrained to stay on the axis of the ring. The electron is then
released from rest.
(a) Describe the subsequent motion of the electron.
(b) Find the speed of the electron when it reaches the centre of the ring.
PHYS 153 08W 13
Problem 23.34
An infinitely long line of charge has linear charge density 5.00x10-12 C/m.
A proton (mass 1.67x10-27 kg. and charge 1.60x10-19 C) is 18.0 cm from
the line and moving directly toward the line at 1.50x103 m/s.
(a) Calculate the proton’s initial K.E.
(b) How close does the proton get to the line of charge?
PHYS 153 08W 14