Physics 6B
Electric Potential and
Electric Potential Energy
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Electric Potential
Measured in Volts
(1Volt  1
Joule
)
Coulomb
Electric Potential Energy
Measured in Joules
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Electric Potential
Measured in Volts
(1Volt  1
Joule
)
Coulomb
Represents the energy it takes to move
exactly 1 Coulomb of charge from one
place to another in an electric field.
Electric Potential Energy
Measured in Joules
Represents the energy it takes to move a
charge from one place to another in an
electric field.
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Electric Potential
Measured in Volts
(1Volt  1
Joule
)
Coulomb
Electric Potential Energy
Measured in Joules
Represents the energy it takes to move
exactly 1 Coulomb of charge from one
place to another in an electric field.
Represents the energy it takes to move a
charge from one place to another in an
electric field.
Formula for potential near point charge Q:
Formula for the potential energy of 2 point
charges Q and q:
kQ
V
r
Uelec 
kQq
r
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Electric Potential
Measured in Volts
(1Volt  1
Joule
)
Coulomb
Electric Potential Energy
Measured in Joules
Represents the energy it takes to move
exactly 1 Coulomb of charge from one
place to another in an electric field.
Represents the energy it takes to move a
charge from one place to another in an
electric field.
Formula for potential near point charge Q:
Formula for the potential energy of 2 point
charges Q and q:
kQ
V
r
Notes:
This is not a vector. Use the sign of the
charge to determine the sign of the
potential.
Potential is defined to be zero when r→∞
We will typically use potential differences
that will look like ΔV. Don’t get voltage
confused with velocity or volume.
Uelec 
kQq
r
Notes:
This is not a vector, so the signs of the
charges may be used in the formula.
Potential Energy is always Potential times
charge:
U
 V q
elec
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Example 1: Two point charges
r
Charges Q and q are separated by distance r.
+q
+Q
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Example 1: Two point charges
r
Charges Q and q are separated by distance r.
+q
+Q
The potential energy of this arrangement is
given by our formula:
kQq
Uelec 
r
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Example 1: Two point charges
r
Charges Q and q are separated by distance r.
+q
+Q
The potential energy of this arrangement is
given by our formula:
kQq
Uelec 
r
This represents the amount of energy it would
take to move these charges to where they are
now, if they started very far apart (r→∞)
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Example 1: Two point charges
r
Charges Q and q are separated by distance r.
+q
+Q
The potential energy of this arrangement is
given by our formula:
kQq
Uelec 
r
This represents the amount of energy it would
take to move these charges to where they are
now, if they started very far apart (r→∞)
Like gravitational potential energy, we only
really care about the difference in potential
energy when the charges move from one
arrangement to another. Our formula defines
zero potential energy – when r→∞.
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Example 1: Two point charges
+q
r
Charges Q and q are separated by distance r.
+Q
The potential energy of this arrangement is
given by our formula:
kQq
Uelec 
r
This represents the amount of energy it would
take to move these charges to where they are
now, if they started very far apart (r→∞)
Like gravitational potential energy, we only
really care about the difference in potential
energy when the charges move from one
arrangement to another. Our formula defines
zero potential energy – when r→∞.
r/3
Now suppose that charge q is moved closer, so
it is a distance r/3 from charge Q.
+q
+Q
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Example 1: Two point charges
+q
r
Charges Q and q are separated by distance r.
+Q
The potential energy of this arrangement is
given by our formula:
kQq
Uelec 
r
This represents the amount of energy it would
take to move these charges to where they are
now, if they started very far apart (r→∞)
Like gravitational potential energy, we only
really care about the difference in potential
energy when the charges move from one
arrangement to another. Our formula defines
zero potential energy – when r→∞.
r/3
Now suppose that charge q is moved closer, so
it is a distance r/3 from charge Q.
+q
+Q
Now the potential energy is larger (it would take
some work to move q closer to Q since they are
the same sign).
How much larger is the energy?
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Example 1: Two point charges
+q
r
Charges Q and q are separated by distance r.
+Q
The potential energy of this arrangement is
given by our formula:
kQq
Uelec 
r
This represents the amount of energy it would
take to move these charges to where they are
now, if they started very far apart (r→∞)
Like gravitational potential energy, we only
really care about the difference in potential
energy when the charges move from one
arrangement to another. Our formula defines
zero potential energy – when r→∞.
r/3
Now suppose that charge q is moved closer, so
it is a distance r/3 from charge Q.
+q
+Q
Now the potential energy is larger (it would take
some work to move q closer to Q since they are
the same sign).
How much larger is the energy?
3 times larger than before
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B
Example 2:
2 charges are initially arranged along a line, as shown.
The following values are given:
q1=+10nC; q2=+5nC; r=10cm.
r
q1
r
r
q2
A
a) What is the electric potential difference (in Volts)
between points A and B in the diagram?
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B
Example 2:
2 charges are initially arranged along a line, as shown.
The following values are given:
q1=+10nC; q2=+5nC; r=10cm.
r
q1
r
r
q2
A
a) What is the electric potential difference (in Volts)
between points A and B in the diagram?
We calculate the potential due to each charge separately, then add
them to get the total potential. At point A we get:
VA 
kq1 kq2

r
r
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B
Example 2:
2 charges are initially arranged along a line, as shown.
The following values are given:
q1=+10nC; q2=+5nC; r=10cm.
r
q1
r
r
q2
A
a) What is the electric potential difference (in Volts)
between points A and B in the diagram?
We calculate the potential due to each charge separately, then add
them to get the total potential. At point A we get:



 9  109 Nm2  10  109 C  9  109 Nm2  5  109 C



kq1 kq2 
C2 
C2 

VA 



r
r
0.1m
0.1m

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B
Example 2:
2 charges are initially arranged along a line, as shown.
The following values are given:
q1=+10nC; q2=+5nC; r=10cm.
r
q1
r
r
q2
A
a) What is the electric potential difference (in Volts)
between points A and B in the diagram?
We calculate the potential due to each charge separately, then add
them to get the total potential. At point A we get:



 9  109 Nm2  10  109 C  9  109 Nm2  5  109 C



kq1 kq2 
C2 
C2 

VA 



r
r
0.1m
0.1m

VA  900 Nm  450 Nm  1350V
C
C
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B
Example 2:
2 charges are initially arranged along a line, as shown.
The following values are given:
q1=+10nC; q2=+5nC; r=10cm.
r
q1
r
r
q2
A
a) What is the electric potential difference (in Volts)
between points A and B in the diagram?
We calculate the potential due to each charge separately, then add
them to get the total potential. At point A we get:



 9  109 Nm2  10  109 C  9  109 Nm2  5  109 C



kq1 kq2 
C2 
C2 

VA 



r
r
0.1m
0.1m

VA  900 Nm  450 Nm  1350V
C
C
B
Similarly at point B we have:
x
VB 
kq1
kq2

2 r
2 r
q1
r
r
r
q2
A
x 2  r 2  r 2  x  2r 2  2  r
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B
Example 2:
2 charges are initially arranged along a line, as shown.
The following values are given:
q1=+10nC; q2=+5nC; r=10cm.
r
q1
r
r
q2
A
a) What is the electric potential difference (in Volts)
between points A and B in the diagram?
We calculate the potential due to each charge separately, then add
them to get the total potential. At point A we get:



 9  109 Nm2  10  109 C  9  109 Nm2  5  109 C



kq1 kq2 
C2 
C2 

VA 



r
r
0.1m
0.1m

VA  900 Nm  450 Nm  1350V
C
C
Similarly at point B we have:
 9  109 Nm2  10  109 C  9  109 Nm2  5  109 C




kq1
kq2
C2 
C2 
VB 



0.14m
0.14m
2 r
2 r

VB  636 Nm  318 Nm  954V
C
C


B

x
q1
r
r
r
q2
A
x 2  r 2  r 2  x  2r 2  2  r
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B
Example 2:
2 charges are initially arranged along a line, as shown.
The following values are given:
q1=+10nC; q2=+5nC; r=10cm.
r
q1
r
r
q2
A
a) What is the electric potential difference (in Volts)
between points A and B in the diagram?
We calculate the potential due to each charge separately, then add
them to get the total potential. At point A we get:



 9  109 Nm2  10  109 C  9  109 Nm2  5  109 C



kq1 kq2 
C2 
C2 

VA 



r
r
0.1m
0.1m

VA  900 Nm  450 Nm  1350V
C
B
C
Similarly at point B we have:
 9  109 Nm2  10  109 C  9  109 Nm2  5  109 C




kq1
kq2
C2 
C2 
VB 



0.14m
0.14m
2 r
2 r

VB  636 Nm  318 Nm  954V
C
C



x
q1
r
r
r
q2
A
x 2  r 2  r 2  x  2r 2  2  r
Thus the potential difference is just 396 Volts (with B at a lower potential than A)
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B
Example 2:
2 charges are initially arranged along a line, as shown.
The following values are given:
q1=+10nC; q2=+5nC; r=10cm.
r
q1
r
q3
r
q2
A
b) Now suppose another charge q3= -4mC moves from
point A to point B. How much work (in Joules) is required to
move the charge?
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B
Example 2:
2 charges are initially arranged along a line, as shown.
The following values are given:
q1=+10nC; q2=+5nC; r=10cm.
r
q1
r
q3
r
q2
A
b) Now suppose another charge q3= -4mC moves from
point A to point B. How much work (in Joules) is required to
move the charge?
We already have the potential difference from part a).
Here is the calculation: VAB  VB  VA  954 V  1350 V  396 V
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B
Example 2:
2 charges are initially arranged along a line, as shown.
The following values are given:
q1=+10nC; q2=+5nC; r=10cm.
r
q1
r
q3
r
q2
A
b) Now suppose another charge q3= -4mC moves from
point A to point B. How much work (in Joules) is required to
move the charge?
We already have the potential difference from part a).
Here is the calculation: VAB  VB  VA  954 V  1350 V  396 V
To get the change in the potential energy, multiply by the amount of charge that is moving:


Uelec  q  VAB   4  103 C   396V   1.6J
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B
Example 2:
2 charges are initially arranged along a line, as shown.
The following values are given:
q1=+10nC; q2=+5nC; r=10cm.
r
q1
r
q3
r
q2
A
b) Now suppose another charge q3= -4mC moves from
point A to point B. How much work (in Joules) is required to
move the charge?
We already have the potential difference from part a).
Here is the calculation: VAB  VB  VA  954 V  1350 V  396 V
To get the change in the potential energy, multiply by the amount of charge that is moving:


Uelec  q  VAB   4  103 C   396V   1.6J
The work done on the system is the same as this change in the potential energy. Another
way to think about it is that the electric field did -1.6J of work, so the potential energy of
the system increased by 1.6J.
Basic rule of thumb:
When the potential energy of the system decreases, positive work is done by the electric force.
When potential energy increases, negative work is done by the electric force (or alternatively,
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positive work is done on the system by outside forces).
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Equipotential Surfaces
The diagrams below show the electric field and some corresponding equipotentials for
various charge distributions. Note that the E-field lines are always perpendicular to
the equipotentials.
If a charge were to move along an equipotential, its potential energy would not
change (all points on an equipotential are at EQUAL potential).
If, instead, a charge were to move from one equipotential to another its potential
energy would change (and there would be a corresponding change in kinetic energy).
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Equipotential Surfaces
Here is a problem for you:
Suppose an electron moves from point A to point B in the diagram below.
What is the change in its potential energy?
What is the change in its kinetic energy?
Could the electron do this spontaneously, or is some outside force required to move it?
A
B
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