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8/10/2019 05 Potential and voltage.pdf
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Electric forces exert on any charge placed in the electric field
Potential and voltage
F
q
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The electric fieldEhas the magnitude and the direction (it is a vector)
F E q=
Given the electric field E, the force exerting on any charge q can be
found as
The direction of the field is taken to be the direction of the force it
would exert on a positive test charge.
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Potential energy in electric field (cont.)
The potential energy of abody in gravitational field:
WG = mgh
F= qE
Ground
F= mgh
m
x
The potential energy of acharge in electric field:
WE = q E x
Electric field gravitational field analogy
Charge q Mass m
Electric field E Gravitational acceleration g
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Mechanical and Electrical potential energy - example
Potential energy of a body in
gravitational field:
WG = mgh
F= qE
Ground
F= mgh
m
x
Potential energy of a charge
in electric field:
WE = q E xA positive charge of 1C was moved by
1mm in the electric field of 10 N/C
against the field lines. What is the change
in the charge potential energy?
A body of the mass 1kg was lifted up by 1
mm in the gravitational field (g= 9.8 m/s2).
What is the change in the body potential
energy?
m=1kg; h=1 mm=10-3 m; g= 9.8 m/s2
WG
= mgh = 1kg 9.8 m/s2 10-3 m=
=9.8 10-3 J; WG increases
q=1C; x=1 mm=10-3 m; E= 10 N/C
WE
= qEx = 1C 10 N/C 10-3 m=
=10 10-3 J; WE increases
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Potential energy in electric field (cont.)
Potential energy of a charge in electric field:
point x1: WE1 = q E x1; point x2: WE2 = q E x2;
A positive charge 1C moved from the point x1 into point x2 separated by 1 mm in the
electric field of 10 N/C in the direction offield lines.
What is the change in the charge potential energy?
q=1C; x=1 mm=10-3 m; E= 10 N/C
WE= WE2 WE1WE = qE (x2 x1) = 1C 10 (-10
-3) = - 10 10-3 J;
Notes: 1) WE decreases the electric field does the work
2) the absolute positions, x1
and x2
do not matter; only
the difference x = x2 x1
F= qE
x1x2
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Electric potential
F= qEx
Potential energy of a charge
in electric field:
WE = q E xA positive charge of 1C was moved by
1mm in the electric field of 10 N/C
against the field lines. What is the change
in the charge potential energy?
q=1C; x=1 mm=10-3 m; E= 10 N/C
WE
= qEx = 1C 10 N/C 10-3 m=
=10 10-3 J. WE increases
Potential is a potential energyof a unit charge in the electric
field; it does not depend on the
charge value: = WE/q [J/C]
=E x [(N/C) m] = [J/C]
Potential is always measured with respectto the reference (zero potential) level.
A charge was lifted by 1mm from the
reference plane in the electric field of 10
N/C against the field lines. What is thecharge potential?
x=1 mm=10-3 m; E= 10 N/C
= Ex = 10 N/C 10-3
m == 10 10-3 Nm/C= 10 10-3 J/C
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Potential is the potential energy of a unit charge in electric field:
= WE/q
Potential is measured in Volts (V)
1V is the potential that changes a potential energy of the unit
charge of 1C (Coulomb) by 1 J (Joule)
Electric potential definition
C
JV
1
11 =
If the potential of any point in the electric field is known, thepotential energy of any charge Q can be found:
WE = Q
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Electric field units: V/m and N/C
=E x; from this
1V = 1 N/C * 1 m;
1 N/C = 1 V/m
[E] = [N/C] = [V/m]
V/m is a commonly accepted unit for the electric field.
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Summary of the electric force, field and potential
concepts Electric forces exist in the space surrounding any charge.
Electric forces exert on any charge located in the vicinity of thesource charges.
The magnitude of electric forces can be characterized by electric
fields.
Electric field is the electric force per unit charge.
The potential energy of a charge in electric field is characterized by
the potential.
Potential (Volts) is the potential energy in electric field per unit
charge.
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Negatively charged plate creates a uniform electric field of 103 V/m.
(a) What is the potential of a point 1 mm above the plate?
Example problem 1
=E x; x = 1 mm = 10-3m
=E x = 1.0103 V/m 10-3m = 1.0 V;
(b) What is the potential of a point that is 2 mm above the plate?
=2.0 V;
(c) What is the potential of a point directly on the plate?
= 0 V
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Negatively charged plate creates a uniform electric field of 103 V/m.
Point 1 is located 1 mm above the plate;
point 2 is located 2 mm above the plate.
What is the difference in the potentials of the two points?
Example problem 2
1 =E x1; x = 1 mm = 10-3m
1 =E x1 = 1.0103
V/m 10-3
m = 1.0 V;
2 =E x2; x = 2 mm = 2 10-3m
2 =E x2 = 1.0103 V/m 2 10-3m = 2.0 V;
2 1 =2.0V 1.0 V = 1 V
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Negatively charged plate creates a uniform electric field of 103 V/m.What is the change in the potential of a charge
that has been moved up by d = 1 mm?
The distance from each of the points 1 and 2 to the plate is unknown.Assume point 1 is d0 mm away from the plate.
For the point 2 the distance would be (d0 + d) as the charge moves UP.
1 =E d0; (d0 unknown)2 =E (d0 + d) ; (d0 unknown)
The change in the potential
= 2222 - 1111 = E (d0+d) - Ed0 = E d
= E d = 103 V/m x 10-3mm = 1 VImportant observation :the potential difference
DOES NOT depend on the absolute position of the starting point.
Example problem 3
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What is the change in the potential energy of the charge Q=1 nCthat has been moved from the point with the potential 2 V to the point
with the potential of 5 V?
The potential energy of a charge in the electric field:
WE = Q
The change in the potential energyWE = Q 2222 - Q 1111 = Q (2222 - 1111))))
WE = 1E-9 C (5V 2V) = 1E-9 C 3V = 3E-9 J
Important observation: the change in the potential energy
depends ONLY on the potential difference.
Example problem 4
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Some conclusions from the above examples
As far as only the changes in potential or potential energy are
concerned, the absolute potentials of the start and end points are not
important: only the difference between them.
(Compare to the mechanical potential energy: only the energychange
is important)
The potential increases as the point moves towards the positive
electrode (AGAINST the field lines) and decreases when it moves
towards the negative electrode (ALONG the filed lines)
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Due to the relative character of potential, the most important energy
characteristic of electric field is the potential difference.
Voltage
The potential difference is also called the voltage V.
Voltage = Potential Difference
Being a potential difference, voltage is also measured in Volts (V)
If the potentials corresponding to the two different points 1 and 2in the electric field are 1 and 2,
the voltage V21 between these points,
V21 = 2 -1
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Electric Potential and Voltage
V21 is the potential energy to move the unit charge from point 1 to point 2:
V21 = 2 - 1
V12 is the potential energy to move the unit charge from point 2 to point 1:
V12 = 1 2
V = 0
x1
V21 = 2 -1 =E (x2-x1)
E
x2
1 = E x1
2 = E x2
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Example problem 5
The voltage between two charged parallel plates is 5 V.
The separation between the plates is d =1 mm
Find the electric field between the plates.
Assume the field between the plates is uniform
+
-
Solution
The electric field between the plates is uniform, hence,V = E . d; E = V/d = 5V / 1mm = 5V / 10-3 m = 5 .103 V/m
Answer: E = 5 .103 V/m = 5 kV/m
The electric field direction is vertically downward.
d E
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Example problem 6
The voltage between two charged parallel plates is 5 V.
What energy is acquired by an electron that is moved from the
bottom plate up to the top plate?
-
+
Solution
The potential energy of the electron on the bottom plate is 0 JThe voltage of the top plate is V = -5V with respect to the bottom plate.
The potential energy of the electron that moves across the voltage V,
WE= q V = -e.V = -1.6 10-19C . (-5 V) = 8 .10-19J
Answer: WE= 8 .10-19J
d
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Uniform electric field between two charged plates is 10 V/cm.
What is the voltage between two points A and B separated by
the distance d = 3 mm?
+
-
Solution
dA
B
The potentials are 1 = Ex1; 2 = Ex2
The voltage = potential difference is V21 = 2-1=E.
(x2-x1) = E d;
E = 10 V/cm = 10 V /(10-2 m) = 103 V/m;
V = 103 V/m . 3 . (10-3 m) = 3 V
Example problem 7
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The voltage between any two points in any electric field is equal to the
potential difference between these two points
Vmn = m - nNote that in the voltage indices normally, the first index is the end
point and the second index is the start point
General Voltage - Electric Potential relationship
Arbitrary electric field Points with
different potentials
1
2
3
E
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Solution: Vmn = m - n
V21 = 2 1 = 9V 3V = 6 V;
V32 = 3 2 = 6.5 V 9V = -2.5 V;
V31 = 3 1 = 6.5V 3V = 3.5 V;
V13 = 1 3 = 3V 6.5V = -3.5 V;
The three nodes, 1,2,3
in the amplifier circuit have the potentials
1 = 3 V; 2 = 9 V; 3 = 6.5 V withrespect to the reference node 0
Question 1: find the voltages V21, V32, V31 and V13
Example problem 8
1
2
3
0
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Solution:
The potential energy of the charge Q at the node 1, W1 = 1Q;
The potential energy of the charge Q at the node 2, W2 = 2Q;
The change in the potential energy when the charge Q moves from the
node 1 to the node 2: W21 = W2 W1 = 2Q -1Q =
(2 -1)Q = (9 V 3 V) 1mC = 6V 1mC = 6 VmC = 6 mJ
The energy is positive, i.e. the work needs to done to move the charge
The three nodes, 1,2,3
in the amplifier circuit have the potentials
1 = 3 V; 2 = 9 V; 3 = 6.5 V with
respect to the reference node 0
Question 2: find the energy required to moving the charge Q = 2 mC
from the node 1 to the node 2.
Example problem 8
1
2
3
0
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Solution:from V21 = 2 1,
2 = 1+V21 = 4.5 V + 4 V = 8.5 V;
In the motor driver circuit fragment on the
left, the node 0 has a zero potential (the
node 0 is grounded).
The potential of the node 1 is 1 = 4.5 V.
The potential of the node 3 is 3 = 9 V.
The voltage V21
= 4 V;
Question 1: Find the potential 2
Example problem 9
21
0
3
Question 2: find the voltages V32 and V31
Solution: V32 = 3 2 = 9V 8.5 V = 0.5 V
V31 = 3 1 = 9V 4.5 V = 4.5 V
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