Electric Potential
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Transcript Electric Potential
Electric Potential
Gravitational Potential Energy
A
GPE =
GPE =
F = mg
B
hA
hB
GPE =
required to raise or lower
the book.
-Where
=
Electric Potential Energy
+
+
+
+
+
+
+
+
+
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A +
ΔEPE =
F = qoE
ΔEPE =
dA
-WE(AB) =
B +
dB
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F = qoE
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•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
to move
away from one another and
to move
them closer together.
-qo
F = kqqo
E
r2
+q
r
Ue = Fr =
Electric Potential Energy and
Work of Point Charges
-qo
rB
-qo
rA
+q
+q
A
B
To change the
energy level from
UA to UB, it requires
work (W).
-W =
–
.
-W =
–
.
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
V
Since EPE
V
SI Units:
/
=1
(
)
The Electric Potential Difference is equal to the
required to move a test charge from infinity to a point in
an
divided by the
of the test charge.
The Electric Potential is the
per unit of
( /
).
Relationship Between Electric Potential
and Distance(point charges)
Consider relationship between V and r.
VB - VA =
What
•As
-
happens to V as rB goes to ?
r increases, i.e., as rB , V
•The
The
=
relationship above reduces to: V =
.
/
of the charge will determine if the
electric potential is
or
.
When two or more charges are present, the total
electric potential is the
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.
V=
+
+
Note that the electric potential
can be determined from any
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
r3
r1 r2
Remember:
=
Q1
Q2
r2
Q3
.
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
Characteristics of a Capacitor
Two equal and
oppositely
charged plates
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
qo
B
E
qo
C
qo
A
-
Uniform Electric
Field
• Since the
is
force acting on a charged particle will be the
everywhere between the plates.
• Fe =
, the
Electric Potential and Work in a
Capacitor
WAB =
-
WAB =
V =
=
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
D
A
B
qo
qo
F = qoE
dB
F = qoE
If WAB = qoEd, then what is WCD?
WCD =
Joules because the
the
of
.
•Do you remember that W =
dA
C
qo
acts
-
to
?
Electric Potential of a
Capacitor – An alternative
From mechanics, W =
.
From the previous slide, W =
From the reference table, V =
Two equal and
oppositely
charged plates
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
V =
A
.
.
B
qo
F = qoE
Uniform
Electric
Field
d
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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?
d
Example 3: Parallel Plates
A proton and an electron are released from rest
from a similarly charged plate 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.)
+
+
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+
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+
+
+
p+
e-
d
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Begin by drawing a picture and listing
what is known:
V=
d=
qe =
Example 3: Parallel
Plates(#1 & #2)
For #1, you could answer #2 first to
verify.
The answer is that the
particles will be the
• Why?
• because of the formula needed in question #2
applies to both charges, and
=
.
• Hence:
of
Example 3: Parallel
Plates(#3)
Apply the
to
determine the final speed of the electron and
proton.
Since the
equal to
Proton:
Electron:
:
is
Example 3: Parallel
Plates(#4)
Since F =
, it will be the
for both
particles because their
are the
and the electric field is
between two parallel plates.
We also know that W =
. Since we know
the
between the
and the
done to move either charge from one
plate to another, we can determine the force as
follows:
Example 3: Parallel
Plates(#5)
Since we have the
acting on each
particle, we can now calculate the
each particle using
.
of
Equipotential Lines
Equipotential lines denote where the electric potential is
the
in an electric field.
The potential is the
anywhere on an
a distance
from a
point charge, or
from a plate.
No
is done to move a charge
an
. Hence
=
(The
electric potential difference does
depend on
the
taken from
to
).
Electric field lines and equipotential lines cross at
and point in the direction of
potential.
Equipotential Lines
Parallel Plate Capacitor
+
+
+
+
+
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+
+
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+
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+
+
+
Electric Potential /
Voltage
-
Note:
Equipotential Lines
Point Charge
Note:
+
Note: A charged
surface is also an
Electric Potential /
Voltage
!
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.
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+
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+
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+
A
B
qo
qo
F = qoE
dB
F = qoE
dA
WAB = EPEB – EPEA
WAB = FdB – FdA
WAB = qoEdB – qoEdA
WAB = qoE(dB – dA) = qoEd
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