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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