Download Electric Potential

Electric Potential
Gravitational Potential Energy
A
GPE = mgΔh
GPE = mghA – mghB
F = mg
B
hA
GPE = Work (W) required
to raise or lower the
book.
•Where W = (Fgravity)(Δh)
hB
F = mg
Electric Potential Energy
W = Fd = (Eq)d
As the + charges are
moved closer, work is
required to effect this
change in potential energy
+
+
Electric Potential Energy
+
+
+
+
+
+
+
+
+
-
-
A +
ΔEPE = qoEΔd
F = qoE
ΔEPE = qoEdA – qoEdB
dA
-WE(AB) = qoEdA – qoEdB
B +
dB
-
-
-
-
-
F = qoE
-
-
-
•Does a proton at rest at point A have more or less
potential energy than it would at point B? More
Electric Potential Energy of
Point Charges



Much like the book is attracted to the earth due to gravity,
two unlike charges are attracted to one another.
Conversely, like charges repel.
It takes positive work to move unlike charges away from
one another and negative work to move them closer
together.
-qo
F = kqqo
E
r2
+q
r
EPE = Fr = kqqo
r
Electric Potential Energy and
Work of Point Charges
-qo
rB
-qo
-W = EPEB – EPEA
rA
-W = kqqo – kqqo
rB
rA
+q
+q
A
To change the
energy level from A
to B, it requires
positive work (W).
B
Electric Potential Energy
1.
What would happen if the charged particle q was fixed in
place and then particle qo was suddenly released from
rest?
A.
It would accelerate away from q.
It would accelerate towards q.
It would stay where it is.
B.
C.
2.
How would the potential energy of this
system change?
A.
It would increase.
It would decrease.
It would remain the same.
B.
C.
-qo
+q
Electric Potential
Potential Difference
V  VB  VA 
EPE  W

qo
qo
kqqo
Since EPE 
r
kq kq
V 

rB rA
SI Units: joule/coulomb = 1 volt (V)


The Electric Potential Difference is equal to the Work
required to move a test charge from infinity to a point in
an electric field divided by the magnitude of the test
charge.
The Electric Potential is the energy per unit of charge
(J/C).
Electric Potential
Voltage is defined as the work done
per unit of charge, or
 V = W/q

SI Units: joule/coulomb = 1 volt (V)


The absolute electric potential is equal to the Work
required to move a test charge from infinity to a point in
an electric field divided by the magnitude of the test
charge.
The Electric Potential is the energy per unit of charge
(J/C).
Relationship Between Electric Potential
and Distance(point charges)

Consider relationship between V and r.
-WAB
VB - VA =
=
qo
What
•As
kq
rA
happens to V as rB goes to ?
r increases, i.e., as rB  , V  0.
•The
The
kq
rB
relationship above reduces to: V = kq/r
sign of the charge will determine if the
electric potential is positive or negative.
When two or more charges are present, the total
electric potential is the sum total from all the
charges present in the system.
Electric Potential(point charges)


Consider the following system of three point charges.
What is the electric potential that these charges give rise
to at some arbitrary point P?
Use superposition to determine V.
kQ1
V= r +
1

kQ2
+
r2
kQ3
r3
Note that the electric potential
can be determined from any
arbitrary point in space.
Q1
Q2
Q3
r2
r1
r3
P
Electric Potential and Electrical
Potential Energy/Work (point charges)

If we now move a test charge from infinity to
point P, we can determine the potential energy
of the system or the work required to the test
charge to its new location.
U  W  qoV
 Q1 Q2 Q3 
W  kqo  
 
 r1 r2 r3 

Remember: Work = Energy.
Q1
Q2
Q3
r2
r1
r3
qo
Example 1: Two Point Charges
Two point charges, +3.00 µC and -6.10 µC, are separated by
1.00 m. What is the electric potential midway between
them?
+3.00 μC
-6.10 μC
B
A
0.5 m
0.5 m
Vtotal = VA + VB = kqA/rA + kqB/rB
VA = (8.99 x 109 Nm2/C2)(-6.10 x 10-6) = -109678 V
0.5m
VB = (8.99 x 109 Nm2/C2)(3.00 x 10-6) = 53,940 V
0.5m
Vtotal = -55700 V
Characteristics of a Capacitor
Two equal and
oppositely
charged plates
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
qo
B
E
qo
C
qo
A
-
Uniform Electric
Field
• Since the electric field is constant, the force acting
on a charged particle will be the same everywhere
between the plates.
• Fe = qoE
FA = FB = FC
Electric Potential and Work in
a Capacitor
WAB = F·dB - F·dA
WAB = qoEd
(EPE) -WAB
V =
qo = qo
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
D
A
B
qo
qo
F = qoE
dB
F = qoE
dA
C
qo
If WAB = qoEd, then what is WCD?
WCD = 0 Joules because the force acts perpendicular to the
direction of motion.
•Do you remember that W = F·d·cos?
-
Electric Potential of a
Capacitor – An alternative



From mechanics, W = Fd.
From the previous slide, W = qoEd
From the reference table, V = W/qo
Two equal and
oppositely
charged plates
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
A
B
qo
F = qoE
Uniform
Electric
Field
d
-
V = WAB/qo = Fd/qo = qoEd/qo = Ed
Example 2:Parallel Plates
A spark plug in an automobile engine consists of two metal
conductors that are separated by a distance of 0.50 mm. When
an electric spark jumps between them, the magnitude of the
electric field is 4.8 x 107 V/m. What is the magnitude of the
potential difference V between the conductors?
V = Ed
V = (4.8 x 107 V/m)(5.0 x 10-4m)
V = 24,000V
d
Example 3: Parallel Plates
A proton and an electron are released from rest
from two opposite but similarly charged plates of a
capacitor. The electric potential is 100,000 V and
the distance between the two plates is 0.10 mm.
1.
2.
3.
4.
5.
Which charge will have greater kinetic energy at the
moment it reaches the opposite plate?
Determine the amount of work done on each particle.
Determine the speed of each particle at the moment it
reaches the opposite plate.
Determine the magnitude of the force acting on each
particle.
Determine the magnitude of the acceleration of each
particle.
Example 3: Parallel
Plates(cont.)
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+

p+
e-
d
-
Begin by drawing a picture and listing
what is known:



V = 100,000V
d = 0.10 mm = 1.0 x 10-4m
qe = qp = 1.6 x 10-19C (ignore the sign. We
are only interested in magnitude.)
Example 3: Parallel
Plates(#1 & #2)

For #1, you could answer #2 first to
verify.

The answer is that the kinetic energy of both
particles will be the same
• Why?
• because of the formula needed in question #2
applies to both charges, and work = energy.
• Hence: Wproton = Welectron
qprotonV = qelectronV
Wproton = Welectron = (1.6x10-19C)(100,000V)
Wproton = Welectron = 1.6x10-14 J
Example 3: Parallel
Plates(#3)


Apply the work-energy theorem to determine the
final speed of the electron and proton.
W = KE
Since the initial kinetic energy is equal to 0J:
W = KEf
W = ½ mvf2
(2)(W )
(2)(1.6 1014 J )
6
vf 


4.38

10
m/s
27
(m proton )
(1.67 10 kg )

Proton:

(2)(W )
(2)(1.6 1014 J )
8


1.87

10
m/ s
Electron: v f 
31
(melectron )
(9.1110 kg )
Example 3: Parallel
Plates(#4)


Since F = qE, it will be the same for both
particles because their charges are the same
and the electric field is uniform between two
parallel plates.
We also know that W = Fd. Since we know the
distance between the plates and the work done
to move either charge from one plate to
another, we can determine the force as follows:
W 1.6 1014 J
10
F 

1.6

10
N
4
d 1.0 10 m
Example 3: Parallel
Plates(#5)

Since we have the force acting on each particle,
we can now calculate the acceleration of each
particle using Newton’s 2nd Law.
a
a
F
m proton
F
melectron
10
1.6 10 N
16
2

 9.6 10 m / s
27
1.67 10 kg
1.6 1010 N
20
2

 1.8 10 m / s
31
9.1110 kg
Equipotential Lines




Equipotential lines denote where the electric potential is
the same in an electric field.
The potential is the same anywhere on an equipotential
surface a distance r from a point charge, or d from a plate.
No work is done to move a charge along an equipotential
surface. Hence VB = VA (The electric potential
difference does not depend on the path taken from A
to B).
Electric field lines and equipotential lines cross at right
angles and point in the direction of decreasing potential.
Equipotential Lines

Parallel Plate Capacitor
Lines of
Equipotential
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
Decreasing Electric
Potential / Voltage
-
Electric Field
Lines
Note: Electric
field lines and
lines of
equipotential
intersect at right
angles.
Equipotential Lines

Point Charge
Note: Electric
field lines and
lines of
equipotential
intersect at right
angles.
Lines of
Equipotential
+
Decreasing Electric
Potential / Voltage
Electric Field
Lines
Note: A charged
surface is also
an equipotential
surface!
Equipotential Lines (Examples)

http://www.cco.caltech.edu/~phys1/java/phys1/
EField/EField.html
Key Ideas





Electric potential energy (U) is the work required to bring
a positive unit charge from infinity to a point in an electric
field.
Electric potential (V) is the change in energy per unit
charge as the charge is brought from one point to another.
The electric field between two charged plates is constant
meaning that the force is constant between them as well.
The electric potential between two points is not dependent
on the path taken to get there.
Electric field lines and lines of equipotential intersect at
right angles.
Electric Potential Energy and Work
in a Uniform Electric Field
Note: The force
acting on the charge
is constant as it
moves from one
plate to another
because the electric
field is uniform.
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
A
B
qo
qo
F = qoE
dB
F = qoE
dA
WAB = EPEB – EPEA
WAB = FdB – FdA
WAB = qoEdB – qoEdA
WAB = qoE(dB – dA) = qoEd
-