Download Notes - Electrostatics

3/5/2009
PHYS202 – SPRING 2009
Lecture notes ‐ Electrostatics
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What is Electricity
• Static effects known since ancient times.
• Static charges can be made by rubbing certain materials together.
• Described by Benjamin Franklin as a fluid.
– Excess of the fluid was described as positive.
– Deficit of the fluid was described as negative.
• Like charges repel, unlike charges attract.
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A Quantitative Approach
• Charles Augustin de Coulomb first described relationship of force and charges.
• Coulomb invented the torsion balance to measure the force between charges.
• Basic unit of charge in SI system is called the Coulomb
• One of the basic irreducible measurements.
– Like mass, length and time.
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Fundamental Charge
•
Due to atomic structure.
•
Positively charged heavy nuclei and negative charged electrons.
•
Charged objects have excess or deficit of electrons.
•
Protons and electrons have equal and opposite
charge: e = 1.602 x 10‐19 C
•
Rubbing some objects causes electrons to
move from one object to another.
•
Negatively charged objects have an excess
of electrons.
•
Positively charged objects have a deficit of
electrons.
•
Net change in charge in a closed system must
be zero.
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Fundamental Charge
Problem 18.6 ‐ Water has a mass per molecule of 18 g/mol and each molecule has 10 electrons. How many electrons are in one liter of water?
10 electrons 6.022×10 23 molecules 1mol 1g 103 ml
×
×
N=
×
×
1molecule
1mol
18g 1ml 1liter
N = 3.3 ×1026
electrons
liter
What is the net charge of the electrons?
(
)(
q = Ne = 3.3 ×1026 electrons −1.602 × 10−19 C / electron
)
q = −5.4 ×107 C
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Conservation of Charge
•
Net change in charge in any closed system is zero.
•
Or, the change of charge in any volume equals the net charge entering and leaving the system:
( Charge ) = ( Original Charge ) + ( Charge in ) − ( Charge out )
•
We denote the amount of charge by the letter Q or q
•
If Charge is created, it must be equal and opposite.
q = q0 + qin - qout
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Conservation of Charge
Two identical metal spheres of charge 5.0 C and ‐1.6 C are brought together and then separated. What is the charge on each sphere?
qtotal = q1 + q2 = 3.4 C
Since the charge is distributed equally, each must have an equal charge:
qtotal
q1 = q2 =
= 1.7 C
2
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Conductors and Insulators
• Insulator
– Charge stays where it is deposited.
– Wood, mica, Teflon, rubber.
• Conductor
– Charge moves “freely” within the object.
– Metals exhibit high conductivity
• Due to atomic structure.
– Is the outermost electron(s) tightly bound?
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Charging Objects
• By friction
– Rubbing different materials
– Example: fur and ebonite
• By contact
– Bringing a charged object into contact with another (charged or uncharged) object.
• By induction
– Bringing a charged object near a conductor temporally causes opposite charge to flow towards the charged object.
– Grounding that conductor “drains” the excess charge leaving the conductor oppositely charged.
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Force Between Charges
• Coulomb found that the force between two charges is proportional to the product of their charges and inversely proportion to the square of the distance between them:
q1 q 2
F ∝ 2
r12
Where q1 and q2 represent the charges and r12 the distance between them.
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Coulomb’s Law
•
By using the torsion balance, Coulomb discovered the constant of proportionality: k = 8.99 x 109 N·m2/C2
q1 q 2
F =k 2
r12
+
+
‐
‐
‐
+
•
If both q1 and q2 are both positive, the force is positive.
•
If both q1 and q2 are both negative, the force is positive.
•
If both q1 is positive and q2 is negative, the force is negative.
•
Hence, the sign of the force is positive for repulsion and negative for attraction.
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Coulomb’s Law
Principle of Superposition
• Electric force obeys the law of linear superposition – the net force on a charge is the vector sum of the forces due to each charge:
G
Fi =
∑k
qi q j
2
ij
r
i≠ j
• Add the vectors:
rˆij
3
1
F34
The net force on charge
4 is F14+F24+F34=Fnet
2
Red are positive charges
Blue are negative charges
F24
4
F14
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Coulomb’s Law Principle of Superposition
• The forces add for as many charges as you have:
G
Fi =
∑k
i≠ j
qi q j
2
ij
r
rˆij
6
1
3
8
5
Red are positive charges
Blue are negative charges
4
2
7
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Coulomb’s Law Principle of Superposition
• Electric force obeys the law of linear superposition – the net force on a charge is the vector sum of the forces due to each charge:
G
Fi =
∑k
i≠ j
qi q j
2
ij
r
6
3
1
8
5
The net force on charge 5 is the
vector sum of all the forces on
charge 5.
Red are positive charges
Blue are negative charges
4
2
rˆij
7
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Coulomb’s Law
18.14 – Two tiny conducting spheres are identical and carry charges of ‐20.0 μC and +50.0 μC. They are separated by a distance of 2.50 cm. (a) What is the magnitude of the force that each sphere experiences, and is the force attractive or repulsive? (
)(
8.99 × 10 9 N ⋅ m 2 / C 2 − 20.0 × 10 − 6 C
q1 q 2
F =k 2 =
2
r12
2.50 × 10 − 2 m
(
)
)( 50.0 × 10
−6
C
)
F = − 1.44 × 10 4 N
The net force is negative so that the force is attractive.
(b) The spheres are brought into contact and then separated to a distance of 2.50 cm. Determine the magnitude of the force that each sphere now experiences, and state whether the force is attractive or repulsive.
Since the spheres are identical, the charge on each after being separated is one‐
half the net charge, so q1 + q2 = +15μC . F =k
q1q2
r122
8.99 × 109 N ⋅ m 2 /C 2 )(15.0 × 10−6 C )(15.0 × 10−6 C )
(
=
=
2
−2
( 2.50 × 10 m )
3.24 × 103 N
The force is positive so it is repulsive.
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Coulomb’s Law
18.21 – An electrically neutral model airplane is flying in a horizontal circle on a 3.0 m guideline, which is nearly parallel to the ground. The line breaks when the kinetic energy of the plane is 50.0 J. Reconsider the same situation, except that now there is a point charge of +q on the plane and a point charge of −q at the other end of the guideline. In this case, the line breaks when the kinetic energy of the plane is 51.8 J. Find the magnitude of the charges.
From the definition of kinetic energy, we see that mv2 = 2(KE), so that Equation 5.3 for the centripetal force becomes
m υ2 2 ( KE )
Fc =
=
r
r
So,
kq
2
Tmax + 2 =
r
Centripetal
force
2 ( KE )charged
r
(1)
Tmax =
Centripetal
force
2 ( KE )uncharged
r
( 2)
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Coulomb’s Law
18.21 – (cont.)
kq
2
Tmax + 2 =
r
2 ( KE )charged
r
Centripetal
force
(1)
Tmax =
2 ( KE )uncharged
Centripetal
force
r
( 2)
Subtracting Equation (2) from Equation (1) eliminates Tmax and gives
kq
r2
2
2 ⎡( KE )charged – ( KE )uncharged ⎤
⎦
= ⎣
r
Solving for |q| gives
2r ⎡( KE )charged – ( KE )uncharged ⎤
⎣
⎦ = 2 ( 3.0 m )( 51.8 J – 50.0 J ) = 3.5 × 10 –5 C
q =
k
8.99 × 109 N ⋅ m 2 / C 2
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The Electric Field
• Electric charges create and exert and experience the electric force.
G
Fi =
∑k
i≠ j
qi q j
2
ij
r
rˆij
G
qj
Fi = q i ∑ k 2 rˆij
rij
i≠ j
⇒
• We define the Electric Field as the force a test charge would feel if placed at that point:
G
E=
qj
∑kr
i≠ j
2
rˆ
⇒
G
G
F = qE
• The electric field exists around all charged objects.
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Electric Force Lines
• Michael Faraday saw the force caused by one charge to be from lines emanating from the charges.
• Lines begin on Positive charges and end on Negative charges, flowing smoothly through space, never crossing.
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Electric Force Lines
• Faraday imagined these lines and the force was greater where the lines are denser, weaker where they are spread out.
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Electric Field Lines
• Go out from positive charges and into negative charges.
Positive Negative Neutral
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The Inverse Square Law
• Many laws based on the geometry of three‐
dimensional space:
I sound
Gm1m2
Fg = −
r2
P
L
I
=
=
light
4π r 2
4π r 2
qq
Fe = k 1 2 2
r
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The Electric Field
18.28 – Four point charges have the same magnitude of 2.4 x 10‐12 C and are fixed to the corners of a square that is 4.0 cm on a side. Three of the charges are positive and one is negative. Determine the magnitude of the net electric field that exists at the center of the square.
+ C
B +
The drawing at the right shows each of the field contributions at the center of the square (see black E
dot). Each is directed along a diagonal of the square. E
Note that ED and EB point in opposite directions and, E
E
therefore, cancel, since they have the same magnitude. In contrast EA and EC point in the same direction toward + D
corner A and, therefore, combine to give a net field that A is twice the magnitude of EA or EC. In other words, the net field at the center of the square is given by the following vector equation:
ΣE = E A + E B + EC + E D
D
A
B
C
= E A + E B + EC − E B
= E A + EC
= 2E A
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The Electric Field
+ C
B +
18.28 – (cont.)
kq
The magnitude of the net field is Σ E = 2 EA = 2 2
r
ED
EA
In this result r is the distance from a corner to the center of the square, which is one half of the diagonal distance d. Using L for the length of a side of the square A
and taking advantage of the Pythagorean theorem, we r = L/ 2
have . With this substitution for r, the magnitude of the net field becomes
ΣE = 2
ΣE =
(
kq
(
L/ 2
)
2
=
EC
-
EB
+ D
4k q
L2
)(
4 8.99 ×109 N ⋅ m 2 / C2 2.4 ×10−12 C
)
( 0.040 m )2
Σ E = 54 N/C
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Electric Field Lines
18.34 – Review Conceptual Example 12 before attempting to work this problem. The magnitude of each of the charges in Figure 18‐21 is 8.6 x 10‐12 C . The lengths of the sides of the rectangles are 3.00 cm and 5.00 cm. Find the magnitude of the electric field at the center of the rectangle in Figures 18‐21a and b. The figure at the right shows the configuration given in text Figure +q
18.21a. The electric field at the center of the rectangle is the resultant 1 1
of the electric fields at the center due to each of the four charges. As discussed in Conceptual Example 11, the magnitudes of the electric field at the center due to each of the four charges are equal. However, the 4 4
fields produced by the charges in corners 1 and 3 are in opposite directions. Since they have the same magnitudes, they combine to give + q
zero resultant. The fields produced by the charges in corners 2 and 4 point in the same direction (toward corner 2). Thus, EC = EC2 + EC4, where EC is the magnitude of the electric field at the center of the rectangle, and EC2 and EC4 are the magnitudes of the electric field at the center due to the charges in corners 2 and 4 respectively. Since both EC2 and EC4 have the same magnitude, we have EC = 2 EC2.
- q
2
3
+q
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Electric Field Lines
18.34 – (cont.) The distance r, from any of the charges to the center of the rectangle, can be found using the Pythagorean theorem:
2
1 1
2
d
5.00 cm
2
d = (3.00 cm) +(5.00 cm) = 5.83 cm
d
Therefore, r = = 2.92 cm = 2.92 × 10−2 m
2
EC = 2 EC 2 =
2kq2
r2
θ
4 4
3
3.00 cm
2(8.99 × 109 N ⋅ m 2 /C2 )(−8.60 × 10−12 C)
2
=
=
−
×
1.81
10
N/C
(2.92 × 10−2 m) 2
The figure at the right shows the configuration given in text Figure 18.21b. All four charges contribute a non‐zero component to the electric field at the center of the rectangle. As discussed in Conceptual Example 11, the contribution from the charges in corners 2 and 4 point toward corner 2 and the contribution from the charges in corners 1 and 3 point toward corner 1. - q
- q
1 1
2
E 13
E 24
C
4 4
+q
3
+q
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Electric Field Lines
18.34 – (cont.) Notice also, the magnitudes of E24 and E13 are equal, and, from the first part of this problem, we know that
E24 = E13 = 1.81 × 102 N/C The electric field at the center of the rectangle is the vector sum of E24 and E13. The x components of E24 and E13 are equal in magnitude and opposite in direction; hence (E13)x – (E24)x = 0
Therefore,
- q
- q
1 1
2
E 13
E 24
C
4 4
+q
3
+q
EC = ( E13 ) y + ( E24 ) y = 2( E13 ) y = 2( E13 ) sin θ
From Figure 2, we have that
sinθ =
5.00 cm 5.00 cm
=
= 0.858
d
5.83 cm
(
)
EC = 2 ( E13 ) sin θ = 2 1.81×102 N/C ( 0.858 ) = 3.11×102 N/C
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The Electric Field
• The electric field exists around all charged objects.
• This is shown by
electric field
lines. • Constructed
identically to the
force lines.
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Drawing Electric Field Lines
• Number of lines at a charge is proportional to the charge.
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Electric Field Lines
• The charge with ‐3q must have three times as many lines as the charge with +q and 1.5 times the lines for +/‐ 2q.
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Parallel Plates of Charge
• Adding fields from each point on surface causes field lines parallel to the plates to cancel.
Positively
Charged
Negatively
Charged
• Away from the edges, the field is parallel and perpendicular to the plate(s).
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Electric Field & Conductors
•
•
•
•
In conductors excess electrons will move when an electric field is present.
For “steady state” (i.e. no moving charges), the net electric field must be zero.
G
G
F = qE
⇒
G
G
F=E=0
Therefore the charges move to the surface(s) of the conductor until they cancel the electric field inside the conductor.
Field lines can start and end on a conductor at right angles to the surface of the conductor.
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Electric Field Lines
18.44 – Two particles are in a uniform electric field whose value is + 2500 N/C. The mass and charge of particle 1 are m1 = 1.4 x 10‐5 kg and q1 = ‐7.0 μC, while the corresponding values for particle 2 are m2 = 2.6 x 10‐5 kg and q2 = +18 μC. Initially the particles are at rest. The particles are both located on the same electric field line, but are separated from each other by a distance d. When released, they accelerate, but always remain at this same distance from each other. Find d.
d
Ex
q1 = −7.0 μC
q2 = +18 μC
m1 = 1.4 × 10−5 kg
m2 = 2.6 × 10−5 kg
+x
The drawing shows the two particles in the electric field Ex. They are separated by a distance d. If the particles are to stay at the same distance from each other after being released, they must have the same acceleration, so ax,1 = ax,2. According to Newton’s second law, the acceleration ax of each particle is equal to the net force ΣFx acting on it divided by its mass m, or .
a x = ΣFx / m
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Electric Field Lines
d
Ex
18.44 – (cont.)
q1 = −7.0 μC
q2 = +18 μC
m1 = 1.4 × 10−5 kg
m2 = 2.6 × 10−5 kg
+x
Note that particle 1 must be to the left of particle 2. If particle 1 were to the right of particle 2, the particles would accelerate in opposite directions and the distance between them would not remain constant.
ΣFx, 1 = q1Ex − k
ax,1 =
ΣFx, 1
m1
q1q2
d2
qq
q1Ex − k 1 22
d
=
m1
ΣFx, 2 = q2 Ex + k
ax, 2 =
ΣFx, 2
m2
q1q2
d2
qq
q2 Ex + k 1 22
d
=
m2
Note the signs indicated the direction of the vector pointing from charge 2 to charge 1 for the force on charge 1.
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Electric Field Lines
d
Ex
18.44 – (cont.)
q1 = −7.0 μC
q2 = +18 μC
m1 = 1.4 × 10−5 kg
m2 = 2.6 × 10−5 kg
+x
The acceleration of each charge must be the same if the distance between the charges remains unchanged:
a1, x = a2, x
q1Ex − k
m1
Solving for d,
d=
q1q2
d2 =
q2 Ex + k
q1q2
d2
m2
kq1q2 ( m1 + m2 )
Ex ( q1m2 − q2 m1 )
= 6.5 m
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Electric Flux
•
•
Flux (from Latin – meaning to flow)
Electric flux measures amount of electric field passing through a surface.
G
E
Flux through surface is Area of surface times the field flowing through that surface.
Φ E = EA
φ
G
E
So, to calculate, find the normal to the surface.
Then find the angle between the field and that normal vector: φ
Φ E = E ( cos φ ) A
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Electric Flux
Φ E = E ( cos φ ) A
This can be expressed as the vector dot product:
G G
ΦE = E ⋅ A
G G
E ⋅ A = E A cos φ
or
G G
E ⋅ A = Ex Ax + E y Ay + Ez Az
G
E
So, to calculate, find the normal to the surface.
Then find the angle between the field and that normal vector: φ
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Electric Flux
18.48 – A rectangular surface (0.16 m x 0.38m) is oriented in a uniform electric field of 580 N/C. What is the maximum possible electric flux through the surface?
The maximum possible flux occurs when the electric field is parallel to the normal of the rectangular surface (that is, when the angle between the direction of the field and the direction of the normal is zero). Then,
Φ E = ( E cos φ ) A = ( 580 N / C ) (cos 0°) ( 0.16 m )( 0.38 m )
Φ E = 35 N⋅ m 2 / C
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Gauss’ Law
•
•
Electric Flux through a closed surface is proportional to the charge enclosed by that surface:
q
Φ E = enc
ε0
The fundamental constant ε0 is called the permittivity of free space. It is related to
Coulomb’s law by k = 1 , ε0 = 8.85 x 10‐12 C2/(N·m2)
4πε 0
•
Gauss Law is most useful when the electric field on the entire surface is constant.
•
Take a point charge. Construct a sphere centered on the point charge (all points on the surface equidistant from the charge) – the electric field is constant.
•
Since the field is radial, it is perpendicular to the concentric spherical surface everywhere so φ = 0, cos φ = 1
ΦE =
•
qenc
ε0
⇒
EA cos φ =
qenc
ε0
The direction of the field is radially outward from the charge (or inward towards the charge if q is negative). q
q
(
)
E 4π r 2 =
enc
ε0
⇒
E=
4πε 0 r 2
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Gauss’ Law for a plate of charge
•
In this case we take our closed surface to be a box oriented such that two faces are parallel to the plate of charge and the others are perpendicular to the plate.
•
Since the electric field from a plate of charge is perpendicular to the surface of the plate, the sides that are perpendicular to the plate have no flux.
•
The charge enclosed by the box is the surface charge density multiplied by the area the box passes through σA
the plate: .
•
The area of the box with flux is 2A:
G
E
ΦE =
q
ε0
⇒
E ( 2 A) =
σA
ε0
⇒
E=
σ
2ε 0
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Gauss’ Law for a line of charge
18.54 – A long, thin, straight wire of length L has a positive charge Q distributed uniformly along it. Use Gauss’ law to find the electric field created by this wire at a radial distance r. As the hint suggests, we will use a cylinder as our Gaussian surface. The electric field must point radially outward from the wire. Because of this symmetry the field must be the same strength pointing radially outward on the surface of the cylinder regardless of the point on the cylinder, thus we are justified in using Gauss’ Law. Long straight wire
Gaussian cylinder
E
Gausssian
cylinder
r
+++ + +
1
+ ++
2
3
L
End view
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Gauss’ Law for a line of charge
18.54 – (cont.) Let us define the flux through the surface. Note that the field is parallel to the endcaps, thus the angle between the normal from the surface (which is parallel to the wire) is perpendicular to the field, thus φ = π/2, so, cos(φ)=0 for the ends. For the side if the cylinder , the flux is
Φ E = EAside = E ( 2π rL )
Long straight wire
Gaussian cylinder
r
+++ + +
Thus, from Gauss’ Law,
ΦE =
qenc
ε0
1
⇒
E=
E ( 2π rL ) =
Q
3
L
ε0
Q
E
2πε 0 rL
It is customary to refer to charge density λ = Q/L ,
+++
2
E=
λ
2πε 0 r
Gausssian
cylinder
End view
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Work
•
•
•
•
•
Let us take an object with charge q in a constant electric field E. If we release this charge, the electric field causes a force to move the charge.
q
F = qE ⇒ a = E
m
Let us define the field as , so there is only one dimension to consider. Let E = Exˆ
us also set the initial velocity and position to be both zero.
Let the object move through a distance d. The equations of motion are
1
υ = at
x = at 2
2
1
So, after a time t, the object moves , x = d d = at 2
⇒
t = 2d / a
2
t = υ/ a
υ
2d
=
a
a
⇒
υ = 2ad
υ= 2
qE
d
m
⇒
1 2
mυ = qEd
2
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Conservation of Energy
1 2
mυ = qEd
2
•
The left hand side is the kinetic energy the particle has after moving a distance d.
•
That means, the right hand side must represent some form of potential energy the object had prior to being released.
•
Types of Energy already covered:
– Translational kinetic energy
– Rotational kinetic energy
– Gravitational potential energy
– Elastic (spring) potential energy
Etotal
•
1 2 1 2
1 2
= mυ + I ω + mgh + kx + ( EPE )
2
2
2
Electric Potential Energy
– Cutnell & Johnson use (EPE), however, it is often more convenient to use U.
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Electric Potential Energy
•
Like gravitational potential energy, EPE is defined with respect to the energy at some location.
•
Importance lies in the difference between electric potential energy between two points.
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Example Problem
19.4 – A particle has a charge of +1.5 μC and moves from point A to point B, a distance of 0.20 m. The particle experiences a constant electric force, and its motion is along the line of action of the force. The difference between the particle’s electric potential energy at A and B is UA – UB = +9.0 x 10‐4 J. Find the magnitude and direction of the electric force that acts on the particle and the magnitude and direction of the electric field that the particle experiences. The work done must be equal to the change in potential energy:
⇒
Work = U A − U B
F Δx = U A − U B
U A − U B 9.0 ×10−4 J
F=
=
= 4.5 ×10−3 N (from A to B)
Δx
( 0.20 m )
But, we know that for a particle in an electric field: F = qE
U −UB
U −UB
qE = A
E= A
⇒
Δx
q Δx
9.0 ×10−4 J
3
3.0
10
N/C
E=
=
×
1.5 ×10−6 C ( 0.20 m )
(
)
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Electric Potential
•
The electric potential energy of an object depends on the charge of an object.
•
Once again, we can find a measurement independent of the objects charge: V =
•
The SI unit for V is measured in Volts ( = Joules/Coulomb)
•
A common unit of energy is the electron Volt.
•
Specifically, How much potential energy does an electron get if moved through a potential of 1 Volt.
( EPE )
q
1eV = 1.602 ×10−19 J
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Electric Potential
•
Like the Electric field, the electric potential is independent of any test charge placed to measure a force.
•
Unlike the electric field, the potential is a scaler, not a vector.
•
By using calculus methods it can be shown that the electric potential due to a point charge is
kq
V=
r
•
Note that the value of the electric potential depends on the sign of the charge.
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Electric Potential
for a point charge
• Example: What was the potential difference due to a charge qA = 10.0 nC, at the distances r0 = 10.0 cm and r1 = 20.0 cm ?
ΔV = V0 − V1 =
kq A kq A
−
r0
r1
⎛1 1⎞
ΔV = kq A ⎜ − ⎟
⎝ r0 r1 ⎠
1 ⎞
⎛ 1
ΔV = 8.99 ×109 N⋅ m 2 / C2 10−5 C ⎜
−
⎟
⎝ 0.1m 0.2 m ⎠
ΔV = 450 V
(
)(
)
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3/5/2009
Electric Potential
for a point charge
• Example: A charge qA is held fixed. A second charge qB is placed a distance r0 away and is released. What is the velocity of qB when it is a distance r1 away?
• We use the conservation of energy:
Energy =
1 2
1 2
1
mυ + U
⇒
mυ0 + U 0 = mυ12 + U1
2
2
2
1 2
mυ1 = U 0 − U1 = qB (V0 − V1 )
2
υ1 =
2q
(V0 − V1 )
m
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3/5/2009
Electric Potential
for a point charge
υ1 =
2q
(V0 − V1 )
m
• Find the electric potential at r0 and r1:
kq
kq
V0 = A
V1 = A
r0
r1
2kq A qB
2q ⎛ kq A kq A ⎞
−
=
υ1 =
⎜
⎟
m ⎝ r0
r1 ⎠
m
⎛1 1⎞
⎜ − ⎟
⎝ r0 r1 ⎠
• Or, υ1 = 2qΔV / m
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3/5/2009
Electric Potential
for a point charge
•
For a single charge, we set the potential at infinity to zero.
V=
•
•
•
•
•
kq
r
So what is the potential energy contained in two charges?
Bring in the charges from infinity.
The first charge is free as it does not experience any forces.
The second moves through the field created by the first.
So, the energy in the system must be
ΔU = q2 ΔV
ΔV = Vr − V∞
kq1 kq1 kq1
ΔV =
−
=
r12
∞
r12
kq1q2
U=
r12
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3/5/2009
Potential Energy
of a collection of charges
• Let’s add a third charge to this group.
ΔU = q2 ΔV1 + q3 ΔV1 + q3 ΔV2
ΔV1 =
kq1
r13
ΔV2 =
kq2
r23
kq1q2 kq1q3 kq2 q3
ΔU =
+
+
r12
r13
r23
• Each successive charge experiences the field (i.e. potential) created by the previous charges. Be careful not to double count!
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3/5/2009
Example Problem
19.13 – An electron and a proton are initially very far apart (effectively an infinite distance apart). They are then brought together to form a hydrogen atom, in which the electron orbits the proton at an average distance of 5.29 x 10‐11 m. What is the change in the electric potential energy?
The first charge comes in for free, the second charge experiences the field of the first, so,
ΔU = q2 ΔV
ΔV =
(
ΔU = −1.602 ×10−19 C
)
ΔV = Vr − V∞
kq1 kq1 kq1
−
=
r12
∞
r12
(8.99 × 10
9
)(
N ⋅ m 2 /C2 1.602 × 10−19 C
)
5.29 × 10−11 m
ΔU = −4.35 ×10−18 J
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3/5/2009
Electric Field and Potential
•
Let a charge be moved by a constant electric field from points A to B (let Δd be that
distance).
W = F Δd
⇒ W = qE Δd
•
Now, we know that the work done must be equal to the negative change in electric potential energy:
−ΔU AB = qE Δd
•
But potential energy is simply:
−qΔVAB = qE Δd
•
Thus:
E=−
ΔV
Δd
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3/5/2009
Electric Field and Potential
E=−
ΔV
Δd
•
This implies that the field lines are perpendicular to lines (surfaces) of equal potential.
•
This equation only gives the component of the electric field along the displacement Δd.
•
To truly find the electric vector field, you must find the components in each direction:
Ex = −
ΔV
Δx
Ey = −
ΔV
Δy
Ez = −
ΔV
Δz
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3/5/2009
Electric Field and Potential
• If the potentials at the following points (x, y) are
V(1,1) = 30 V, V(0.9,1)=15V, and V(1,0.9)=40 V. What is the (approximate) electric field at (1,1)?
Ex = −
Ex = −
ΔV
Δx
Ey = −
30 V − 15 V
= −150 V / m
1.0 m− 0.9 m
E =
( −150 V ) + (100 V )
2
ΔV
Δy
Ey = −
2
Ez = −
ΔV
Δz
30 V − 40 V
= 100 V / m
1.0 m− 0.9 m
= 180 V
⎛ 100 V ⎞
D
=
−
θ E = tan ⎜
34
⎟
⎝ −150 V ⎠
−1
E
y
(1,1)
x
So, 34˚ clockwise of –x axis or 146˚ ccw of +x axis
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3/5/2009
Example
Problem 19.34 – Determine the electric field between each of the points.
E=−
ΔV
Δx
E AB = −
5.0 V − 5.0 V
= 0 V/m
0.20 m− 0.0 m
EBC = −
3.0 V − 5.0 V
= 10.0 V / m
0.40 m− 0.20 m
ECD = −
1.0 V − 3.0 V
= 5.0 V / m
0.80 m− 0.40 m
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3/5/2009
Equipotential Surfaces
•
Electric field lines are perpendicular to equipotential surfaces.
•
Electric field lines are perpendicular to surfaces of conductors.
•
Thus, surfaces of conductors are equipotential surfaces.
•
The interior of conductors have no electric field and are therefore equipotential.
•
The Electric potential of a conductor is constant throughout the conductor.
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3/5/2009
Example
19.68 – An electron is released at the negative plate of a parallel plate capacitor and accelerates to the positive plate (see the drawing), (a) As the electron gains kinetic energy, does its electric potential energy increase or decrease? Why? The electric potential energy decreases. The electric force F is a conservative force, so the total energy (kinetic energy plus electric potential energy) remains constant as the electron moves across the capacitor. Thus, as the electron accelerates and its kinetic energy increases, its electric potential energy decreases. (b) The difference in the electron’s electric potential energy between the positive and negative plates is Upos – Uneg . How is this difference related to the charge on the electron (−e) and to the difference in the electric potential between the plates?
The change in the electron’s electric potential energy is equal to the charge on the electron (–e) times the potential difference between the plates. (c) How is the potential difference related to the electric field within the capacitor and the displacement of the positive plate relative to the negative plate?
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3/5/2009
Example
19.68 – (cont.) (c) How is the potential difference related to the electric field within the capacitor and the displacement of the positive plate relative to the negative plate?
The electric field E is related to the potential difference between the plates and the displacement Δs by E=−
(Vpos − Vneg )
Δs
Note that (Vpos – Vneg) and Δs are positive numbers, so the electric field is a negative number, denoting that it points to the left in the drawing.
Problem The plates of a parallel plate capacitor are separated by a distance of 1.2 cm, and the electric field within the capacitor has a magnitude of 2.1 x 106 V/m. An electron starts from rest at the negative plate and accelerates to the positive plate. What is the kinetic energy of the electron just as the electron reaches the positive plate?
Use conservation of energy:
KE pos + U pos = KEneg + U neg
Total energy at
positive plate
⇒
KE pos + eV pos = KEneg + eVneg
Total energy at
negative plate
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3/5/2009
Example
19.68 – (cont.)
KE pos + eV pos = KEneg + eVneg
But the electron starts from rest on the positive plate, so, the kinetic energy at the positive plate is zero.
eV pos = KEneg + eVneg
KEneg = eV pos − eVneg
(
KEneg = e V pos − Vneg
(V
−V
)
)
pos
neg
But, , so,
E=−
Δs
(
)(
)
KEneg = e ( − E Δs ) = −1.602 ×10−19 C −2.16 ×106 V / m ( 0.012 m )
KEneg = 4.0 ×10−15 J
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3/5/2009
Lines of Equal Potential
• Lines of equal elevation on a
map show equal potential
energy (mgh) .
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3/5/2009
Example
19.29 – Two equipotential surfaces surround a +1.5 x 10‐8 C point charge. How far is the 190‐V surface from the 75.0‐V surface?
The electric potential V at a distance r from a point charge q is V = kq/r. The potential is the same at all points on a spherical surface whose distance from the charge is r =kq/V. The radial distance r75 from the charge to the 75.0‐V equipotential surface is r75 = kq/V75, and the distance to the 190‐V equipotential
surface is r190 = kq/V190. The distance between these two surfaces is
r75 − r190 =
r75 − r190
⎛ 1
kq
kq
1 ⎞
−
= kq ⎜
−
⎟⎟
⎜V
V75 V190
V
190 ⎠
⎝ 75
2
⎛
1 ⎞
⎛ 1
−8
9 N⋅m ⎞
= ⎜⎜ 8.99 ×10
−
⎟ +1.50 ×10 C ⎜
⎟ = 1.1 m
2 ⎟
75.0
V
190
V
C
⎝
⎠
⎝
⎠
(
)
64
3/5/2009
Capacitors
•
A capacitor consists of two conductors at different potentials being used to store charge.
•
Commonly, these take the shape of parallel plates, concentric spheres or concentric cylinders, but any conductors will do.
•
q = CV, where C is the capacitance, measured in SI units of Farads (Coulomb / Volt or C2/ J ), V is the voltage difference between the conductors and q is the charge on each conductor (one is positive, the other negative).
•
Capacitance is depends only upon the geometry of the conductors.
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3/5/2009
Parallel Plate Capacitor
• Field in a parallel plate capacitor is E =
q
ε0 A
• But, the a constant electric field can be defined in terms of electric potential:
V
E=−
d
⇒
V = Ed
• Substituting into the equation for capacitance,
q = CV = CEd
⎛ q ⎞
q =C⎜
⎟d
⎝ ε0 A ⎠
⇒
C=
ε0 A
d
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3/5/2009
Other Capacitors
• Two concentric spherical shells
C=
4πε 0
⎛1 1⎞
⎜ − ⎟
⎝a b⎠
• Cylindrical capacitor has a capacitance per
unit length:
2πε 0
C
=
L
⎛b⎞
ln ⎜ ⎟
⎝a⎠
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3/5/2009
Energy stored in a capacitor
• A battery connected to two plates of a capacitor does work charging the capacitor.
• Each time a small amount of charge is added to the capacitor, the charge increases, so the voltage between the plates increases.
V
V =q/c
q/c
Δq
q
q
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3/5/2009
Energy stored in a capacitor
•
Potential Energy is U = qV
•
We see that adding up each small increase in charge Δq x V is just the area of the 1
triangle.
U=
Such that,
2
q = CV
qV
q2
U=
2V
1
U = CV 2
2
So the energy stored in ANY electric field can be calculated
V
V =q/c
q/c
Δq
q
q
1 ⎛ κε A ⎞
2
U = ⎜ 0 ⎟ ( Ed )
2⎝ d ⎠
1
U = κε 0 E 2 ( Ad )
2
Ad is the volume of the capacitor! So the energy density is: U
1
= κε 0 E 2
u=
Ad
2
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3/5/2009
Example
• How much energy is stored in a parallel plate capacitor made of 10.0 cm diameter circles kept 2.0 mm apart when a potential difference of 120 V is applied?
• Find the capacitance:
C=
ε0 A
d
=
• The energy stored is ε πr V
1
U = CV 2 = 0
2
2d
2
2
8.85 ⋅10
(
=
−12
ε 0π r 2
d
)
C 2 / (N⋅ m 2 ) π ( 0.05 m ) (120 V )
2
2
2 ( 0.002 m )
U = 2.09 ×10−9 J
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3/5/2009
Electric Polarization in Matter
• Electric charge distribution in some molecules is not uniform.
• Example: Water – both H
atoms are on one side.
• In the presence of an
Electric Field, water will
align opposite to the field.
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3/5/2009
Electric Polarization in Matter
• In the absence of a field. Molecules will be oriented randomly.
• In an electric field, they line up opposite to the field.
72
3/5/2009
Electric Polarization in Matter
•
The resulting electric field is the sum of the external field and the fields generated by the molecules. •
The electric field strength through the matter is reduced.
•
The reduction of the field depends on the energy in the molecules, the strength of the field each one makes and the density of the molecules.
•
This can be measured and the material is said to have a dielectric constant ‐ κ.
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3/5/2009
Dielectrics
•
This dielectric constant changes the value of the permittivity in space (ε)
•
Coulombs law in a dielectric is F =
•
Where ε is related to ε0 by a constant κ:
•
The dielectric constant for a vacuum is κ = 1.
•
All other materials have a dielectric constant value greater than 1.
•
•
Water at about 20˚C has κ = 80.4.
Air at about 20˚C has κ = 1.00054.
q1q2
rˆ
2
4πε r12
ε = κε 0
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3/5/2009
Dielectrics
ε = κε 0
• How does this affect the capacitance of parallel plates?
Cd =
εA
=
κε 0 A
d
d
Cd = κ C0
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3/5/2009
Dielectric Breakdown
• If the field in the dielectric becomes too strong (i.e. voltage is too great), the material begins to conduct current – this is called dielectric breakdown.
• When this happens the charge flows through the material to the other plate and neutralizes the capacitor.
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3/5/2009
Example
19.57 – Two hollow metal spheres are concentric with each other. The inner sphere has a radius of 0.1500 m and a potential of 85.0 V. The radius of the outer sphere is 0.1520 m and its potential is 82.0 V. If the region between the spheres is filled with Teflon, find the electric energy contained in this space.
The electric energy stored in the region between the metal spheres is
1
u = κε 0 E 2
2
⇒
1
U = κε 0 E 2 (Volume)
2
The volume between the spheres is the difference in the volumes of the spheres:
4
4
4
(Volume) = π r23 − π r23 = π r23 − r23
3
3
3
(
)
Since E = ΔV/Δs and Δs = r2 –r1, so, E = ΔV/(r2 –r1)
2
⎛ ΔV ⎞ ⎡ 4
1
3
3 ⎤
−8
U = κε 0 ⎜
r
r
π
−
⎟ ⎢
2
2 ⎥ = 1.2 × 10 J
2
⎦
⎝ r2 − r1 ⎠ ⎣ 3
(
)
77