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ELECTRIC POTENTIAL September 19, 2008 Picture a Region of space Where there is an Electric Field...
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Transcript of ELECTRIC POTENTIAL September 19, 2008 Picture a Region of space Where there is an Electric Field...
ELECTRIC POTENTIALELECTRIC POTENTIAL
September 19, 2008September 19, 2008
Picture a Region ofPicture a Region ofspace Where there is an space Where there is an
Electric FieldElectric Field
Imagine there is a particle of charge Imagine there is a particle of charge q at some location.q at some location.
Imagine that the particle must be Imagine that the particle must be moved to another spot within the moved to another spot within the field.field.
WorkWork must be done in order to must be done in order to accomplish this.accomplish this.
Electric Potential
We will be dealing with Work Energy & Conservation
Work must be done to move a charge in an electric field. Let’s do a demo ….
I need some help.
What we will do ….
For the moment, assume the charge has MASS. (It may not.) Assume the charge is initially stationary. The charge is to be moved to the left. The charge is to be moved at CONSTANT velocity.
+charge
E
Mr. ExternalMrs. Fields
Start and Sop
ENERGY is required to bring the charge up to speed (if it has mass).
ENERGY is required to bring the particle back to rest (if it has mass).
The sum of these two is ZERO.
Clearly
Both are doing work. BOTH are applying a force through a
distance. BOTH get tired!
Each does the negative amount of work than the other does.
So, when we move a charge in an Electric Field ..
Move the charge at constant velocity so it is in mechanical equilibrium all the time.
Ignore the acceleration at the beginning because you have to do the same amount of negative work to stop it when you get there.
Summary--Summary--
When an object is moved from one point to another in an Electric Field, It takes energy (work) to move it. This work can be done by an external force (you).
You can also think of this as the FIELDFIELD
doing the negativenegative of this amount of work on the particle.
And also remember:
The net work done by a conservative (field)force on a particle moving
around a closed path is
ZERO!
A nice landscape
mg
h
Work done by external force = mgh
How much work here by gravitational field?
The gravitational case:
Someone else’s path
IMPORTANT
The work necessary for an external agent to move a charge from an initial point to a final point is INDEPENDENT OF THE PATH CHOSEN!
The Electric Field
Is a conservative field. No frictional losses, etc.
Is created by charges. When one (external agent) moves a test
charge from one point in a field to another, the external agent must do work.
This work is equal to the increase in potential energy of the charge.
It is also the NEGATIVE of the work done BY THE FIELD in moving the charge from the same points.
A few things to remember…
A conservative force is NOT a Republican. An External Agent is NOT 007.
Electric Potential EnergyElectric Potential Energy
When an electrostatic force acts between two or more charged particles, we can assign an ELECTRIC POTENTIAL ENERGY U to the system.
The change in potential energy of a charge is the amount of work that is done by an external force in moving the charge from its initial position to its new position.
It is the negative of the work done by the FIELD in moving the particle from the initial to the final position.
Definition – Potential Energy
PE or U is the work done by an external agent in moving a charge from a REFERENCE POSITION to a different position.
A Reference ZERO is placed at the most convenient position Like the ground level in many gravitational
potential energy problems.
Example:
E
Zero Levelq
F
d
Work by External Agent
Wexternal = F d = qEd= U
Work done by the Fieldis:Wfield= -qEd = -Wexternal
A uniform electric field of magnitude 290 V/m is directed in the positive x direction. A +13.0 µC charge moves from the origin to the point (x, y) = (20.0 cm, 50.0 cm).(a) What is the change in the potential energy of the charge field system?[-0.000754] J
AN IMPORTANT DEFINITION
Just as the ELECTRIC FIELD was defined as the FORCE per UNIT CHARGE:
We define ELECTRICAL POTENTIAL as the POTENTIAL ENERGY PER UNIT CHARGE:
q
FE
q
UV
VECTOR
SCALAR
UNITS OF POTENTIAL
VOLTCoulomb
Joules
q
UV
Let’s move a charge from one point to another via an external force. The external force does
work on the particle. The ELECTRIC FIELD
also does work on the particle.
We move the particle from point i to point f.
The change in kinetic energy is equal to the work done by the applied forces. Assume this is zero for now.
q
WV
WUUU
also
WW
K
if
WWKKK
applied
appliedif
fieldapplied
fieldappliedif
0
Furthermore…
VqW
so
q
W
q
UV
applied
applied
If we move a particle through a potential difference of V, the work from an external
“person” necessary to do this is qV
Example
Electric Field = 2 N/C
1 C d= 100 meters
Joules
mCN4102
100)/(2C1qEdPE
Energy. potentialin Change
agent EXTERNALby doneWork
One Step More
Joules
mCN4102
100)/(2C1qEdPE
Energy. potentialin Change
agent EXTERNALby doneWork
Volts 200200101
102
q
PE POTENTIALin Change
6
4
C
J
C
JoulesV
Consider Two Plates
OOPS …
Look at the path issue
The difference in potential between the accelerating plates in the electron gun of a TV picture tube is about 25 000 V. If the distance between these plates is 1.50 cm, what is the magnitude of the uniform electric field in this region?
An ion accelerated through a potential difference of 115 V experiences an increase in kinetic energy of 7.37 × 10–17 J. Calculate the charge on the ion.
ImportantImportant
We defined an absolute level of potential. To do this, we needed to define a
REFERENCE or ZERO level for potential. For a uniform field, it didn’t matter where we
placed the reference. For POINT CHARGES, we will see shortly
that we must place the level at infinity or the math gets very messy!
An Equipotential Surface is defined as a surface on which the potential is constant.
0VIt takes NO work to move a charged particlebetween two points at the same potential.
The locus of all possible points that require NO WORK to move the charge to is actually a surface.
Example: A Set of Equipotenital Surfaces
Back To YesteryearBack To Yesteryear
Field Lines and Equipotentials
EquipotentialSurface
ElectricField
Components
EquipotentialSurface
ElectricField
Enormal
Eparallel
x
Work to move a charge a distancex along the equipotential surfaceIs Q x Eparallel X x
BUT
This an EQUIPOTENTIAL Surface No work is needed since V=0 for such a
surface. Consequently Eparallel=0 E must be perpendicular to the equipotential
surface
Therefore
E
E
E
V=constant
Field Lines are Perpendicular to the Equipotential Lines
Equipotential
)(0 ifexternal VVqWork
Consider Two EquipotentialSurfaces – Close together
V
V+dV
dsab
Work to move a charge q from a to b:
VVectords
dVE
and
dVEds
qdVVdVVqdW
also
qEdsdsFdW
external
appliedexternal
E...
)(E
Where
zyx
kji
Over a certain region of space, the electric potential is V = 5x – 3x2y + 2yz2. Find the expressions for the x, y, and z components of the electric field over this region. What is the magnitude of the field at the point P that has coordinates (1, 0, –2) m?
Typical Situation
dF W
Keep in Mind
Force and Displacement are VECTORS!
Potential is a SCALAR.
UNITS
1 VOLT = 1 Joule/Coulomb For the electric field, the units of N/C can be
converted to: 1 (N/C) = 1 (N/C) x 1(V/(J/C)) x 1J/(1 NM) Or
1 N/C = 1 V/m So an acceptable unit for the electric field is now
Volts/meter. N/C is still correct as well.
In Atomic Physics
It is sometimes useful to define an energy in eV or electron volts.
One eV is the additional energy that an proton charge would get if it were accelerated through a potential difference of one volt.
1 eV = e x 1V = (1.6 x 10-19C) x 1(J/C) = 1.6 x 10-19 Joules.
Nothing mysterious.
Coulomb Stuff: A NEW REFERENCE: INFINITY
204
1
r
qE
Consider a unit charge (+) being brought from infinity to a distance r from a Charge q:
q r
To move a unit test charge from infinity to the point at a distance r from the charge q, the external force must do an amount of work that we now can calculate.
x
Just Do It!
OK, doing it!
AB
r
r
unit
AB
rrkq
r
drkq
dr
qk
dVV
B
A
112
2sr
sE
Set the REFERENCE LEVEL OF POTENTIALat INFINITY so (1/rA)=0.
For point charges
i i
i
r
qV
04
1
For a DISTRIBUTION of charge:
volume r
dqkV
Ponder –
What is the potential a distance d from an infinite plane whose charge per unit area is ?
Given two 2.00-μC charges, as shown in Figure P25.16, and a positive test charge q = 1.28 × 10–18 C at the origin, (a) what is the net force exerted by the two 2.00-μC charges on the test charge q? (b) What is the electric field at the origin due to the two 2.00-μC charges? (c) What is the electrical potential at the origin due to the two 2.00-μC charges?
The three charges in Figure P25.19 are at the vertices of an isosceles triangle. Calculate the electric potential at the midpoint of the base, taking q = 7.00 μC.
A disk of radius R has a non-uniform surface charge density σ = Cr, where C is a constant and r is measured from the center of the disk. Find (by direct integration) the potential at P.
Example: Find potential at Pq1 q2
q3 q4
d
rP
md
r
md
qqqqr
V
919.02
3.1
)(1
4
14321
0
q1=12nC q2=-24nC q3=31nC q4=17nC q=36 x 10-9C
V=350 Volts (check the arithmetic!!)
An Examplefinite line of charge
d
r
x
dx
d
xLLV
and
xd
dxV
xd
dxdV
L
2/122
0
02/122
0
2/1220
)(ln
4
1
)(4
1
)(4
1
P
At P Using table of integrals
What about a rodthat goes from –L to +L??
Example
zR
220
22
0
22
0
12
2
2
Rz
zE
zRzdz
d
z
VE
zRzV
z
z
Which was the result we obtained earlier
disk=charge per unit area
In the figure, point P is at the center of the rectangle. With V = 0 at infinity, what is the net electric potential in terms of q/d at P due to the six charged particles?
Continuing
dd
s
ddd
dds
12.152
4
5
42
222
222
s
1 2 3
45 6
d
qxV
d
qk
d
qk
d
q
d
qkV
d
qqqq
d
qqk
r
qkV
i i
i
91035.8
93.093.8812.1
108
12.1
5533
2/
22
HRW gets 8.49 … one of us is right!
Derive an expression in terms of q2/a for the work required to set up the four-charge configuration in the figure, assuming the charges are initially infinitely far apart.
1 2
3 4
aadiagonal 71.12