Transcript electric field - UCSB Campus Learning Assistance Services

Physics 6B
Electric Potential and
Electric Potential Energy
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Electric Potential
Measured in Volts
(1Volt  1
Joule
)
Coulomb
Electric Potential Energy
Measured in Joules
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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.
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Example 1: Two point charges
r
Charges Q and q are separated by distance r.
+q
+Q
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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→∞)
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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→∞.
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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?
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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?
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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

Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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)
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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?
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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,
Prepared by Vince Zaccone
positive work is done on the system by outside forces).
For Campus Learning
Assistance Services at UCSB