Download Prep 4-5 - Oregon State University

Oregon State University
PH 213
Spring Term 2017
Prep 4-5
Suggested finish date: Friday, May 5
The formats (type, length, scope) of these Prep problems have been purposely created
to closely parallel those of a typical exam (indeed, these problems have been taken from
past exams). To get an idea of how best to approach various problem types (there are
three basic types), refer to these sample problems.
1. Evaluate each statement (T/F/N). As always you must fully and correctly explain your answer and reasoning.
a.
A scalar quantity can be negative.
True. Electric potential energy, UE, and voltage, V (also called electric potential), are both scalar quantities, but
they can be negative, depending on where the zero point has been defined. This is because, like many other scalar
quantities (mass, money), the change in the quantity (i.e. the difference, D) is the quantity of interest. Setting the
zero reference point is merely a way to be able to assign relative values; their difference is what’s important.
b. A positive charge can have negative electric potential energy.
True. UE = qV, so if the positive charge, q, is located at a point of negative voltage, V, its UE will be negative.
c. Any description of a property of empty space is a vector description.
False. Electric potential (“voltage”) is a field—a property of empty space—but it is a scalar property.
d. Voltage is a scalar field.
True. Voltage (electric potential) is a measure of electric potential energy (a scalar) that would be stored by
placing 1 Coulomb of charge at a given point in space.
e. Voltage can be defined as zero at any point desired.
True. See item a, above.
f. C·V have units of energy.
True. Volts (V) are J/C, so C(J/C) would be units of Joules (energy).
g. J·s have units of energy.
False. Joules (J) are already units of energy.
h. The units of V·C are equivalent to units of N/m.
False. V·C has units of energy (J), which are equivalent to N·m, not N/m. (Units of N/m would apply to
the stiffness constant, k, for an ideal spring.)
2. Evaluate each statement (T/F/N). As always you must fully and correctly explain your answer and reasoning.
a. The electric potential energy of a proton is the same at any point in a uniform electric field.
False. UE = qV, and V is not constant at all points in a uniform electric (only along equipotential paths—
paths perpendicular to the field).
b. A proton “falling” in an electric field is decreasing its UE.
True. By “falling” here, we mean the motion of a charge—released from rest to move freely—solely under the
influence of the E-field, which is always to convert potential energy to kinetic energy.
c. An electron “falls” toward higher electric potential energy.
False. By “falling” here, we mean the motion of a charge—released from rest to move freely—solely under the
influence of the E-field, which is always to convert potential energy to kinetic energy.
d. If an electron moves in the direction of an electric field, it is moving toward lower voltage.
True. We have arbitrarily defined the direction of an E-field as being the direction of decreasing voltage.
e. In an electric field, an electron will “fall” toward higher voltage.
True. We have arbitrarily defined the direction of an E-field as being the direction of decreasing voltage,
and an electron “falls” against the direction of the field.
f. An electron moves to higher voltage as it loses electric potential energy.
True. We have arbitrarily defined the direction of an E-field as being the direction of decreasing voltage,
and an electron “falls” (loses UE as it moves) against the direction of the field.
g. If an electron within an electric field is moving in the direction of the field lines, it is increasing its UE.
True. Positive external work must be done on the electron to force it to move in the direction of the field.
2
3. Evaluate each statement (T/F/N). As always you must fully and correctly explain your answer and reasoning.
a. No electric force component acts along the path of a charge as it is moved along an equipotential surface.
True. If this were not true, then work would be required to move the charge along that path, thus changing
its potential energy, which—by definition of an equipotential path—does not happen.
b.
It requires zero force to hold a charge on an equipotential path.
Not enough information. It may be that there is zero E-field in the region surrounding the charge’s position, in
which case the statement would be true. However, there could instead be a non-zero E-field acting perpendicular
to the equipotential path, in which case, unless a force is applied in the opposite direction (also perpendicular to
the path), the charge will “fall” off the path.
c. It requires zero work to move a charge along an equipotential surface.
True. If this were not true—if work were required to move the charge along that path, that would be changing
its potential energy, which—by definition of an equipotential path—does not happen.
d. Equipotential surfaces are always perpendicular to E.
True. If this were not true, then there would be a non-zero component of E directed along the equipotential path
—by definition not possible (see item a, above).
e. In any electric field, if you are moving along an equipotential path (a “voltage contour line”), you must turn
90° to your right in order to move in the direction of the field.
False. There are an infinite number of directions that are perpendicular to any particular direction
f.
g.
Defining V∞ = 0 as is conventional for point charges, it is possible that an empty point located near some
point charges could have both of the following field values: Etotal = 0 and
Vtotal = 0
True. There are many charge configurations that could produce this result. Consider, as one simple example,
an empty point located at the origin, with two protons, one each located at (1, 0) and (–1, 0); and also with two
electrons, one each located at (0, 1) and (0, –1).
If V is constant throughout a region of space, then the magnitude of E must be a non-zero constant in that
region.
False. If V is constant in any direction from a point in that region, then every direction must be an equipotential
path, so there cannot be any component of any E-field along any path. So E must be zero everywhere in that
region.
3
4. a.
b.
c.
The drawing shows 3 different cases for the Case 1:
electric potentials of two points, A and B.
In each case, the same positive charge is Case 2:
moved from A to B. In which case, if
any, is the most work done on the positive
Case 3:
charge by the electric force? Explain fully. VA = 150V
•
VA = 25V
VB = 25V
•
•
VA = 10V
•
VB = 100V
•
VB = 60V
•
Case 1. WE.AB = –Wext = –DUE.AB = –qDVAB
The electric force is exerted by the electric field, whose direction on a positive charge, q, is the direction of
decreasing voltage. So in moving a positive charge from A to B (a net voltage loss—i.e. DVAB is negative—only
in case 1), the force is doing positive work only in case 1.
The force is doing zero work in case 2 (when DVAB is zero).
The force is doing negative work in case 3 (when DVAB is positive).
Point A has a voltage of 6.0 V; point B has a voltage of 12 V. At which of those two points would an electron
have a greater electric potential energy (UE), if it were placed there?
Point A. UE.A = qVA = –6e
UE.B = qVB = –12e
–6e > –12e
Initially, three protons are stationary and located at the vertices of an equilateral triangle, one at each
vertex. Each side of the triangle has a length 10L. An outside agent then pushes the protons closer
together. Afterward, the three protons are again stationary and located at the vertices of a smaller
equilateral triangle, one at each vertex. Each side of the smaller triangle has a length L. How much
energy was added to the system during this process—expressed in terms of k, L and/or e?
For the three-particle system:
That is:
UE.i + Wext = UE.f
Wext = UE.f – UE.i = DUE = (UE12.f + UE13.f + UE23.f) – (UE12.i + UE13.i + UE23.i)
= (ke2/L + ke2/L + ke2/L) – [ke2/(10L) + ke2/(10L) + ke2/(10L)]
= 3ke2/L – 3ke2/(10L) = 2.7ke2/L
d. Two electrons are fixed 2.0 cm apart. A third electron is shot from infinity and comes to rest midway
between the first two. What was the initial speed of that third electron?
Call the two fixed electrons q1 and q2; the third electron is q3.
For the 3-particle system:
Simplifying:
That is:
Thus: UE.i + KT.i = UE.f
UE12.i + UE13.i + UE23.i + KT3.i = UE12.f + UE13.f + UE23.f
But:
KT3.i = UE13.f + UE23.f
UE12.i = UE12.f And:
UE13.i = UE23.i = 0
(1/2)mevi2 = ke2/(.01) + ke2/(.01)
vi = √{4ke2/[(.01)(me)]} = √{(4)(8.99 x 109)(1.60 x 10-19)2/[(.01)(9.11 x 10-31)]} = 318 m/s
e. Textbook problem 28.46, page 835.
For the entire collection of charge:
UE.i + KT.i = UE.f + KT.f
Assigning V = 0 at the tumor: QVi = KT.f
Thus: Q = KT.f/Vi = (0.10)/(10,000,000) = 1.00 x 10-8 C (= 10.0 nC)
(3rd edition only)
That is:
UE.i – UE.f = KT.f – KT.i
f. A proton is fixed at a certain point in space (not allowed to move). An electron (mass = me), initially held at
rest at a distance d from the proton, is released (allowed to move freely). How fast is the electron moving
when the voltage value of its position is three times as much as at its initial position?
For the two particle system:
UE.i = UE.f + KT.f
Thus: vi = √[4ke2/(d·me)]
2
2
–ke /d – (–3ke /d) =
(1/2)mevi2
That is: UE.i – UE.f = KT.f
That is: 2ke2/d = (1/2)mevi2
4
5. a. The figure at right shows three charges.
(i) Find the electric potential at the lower right
hand corner due to the top two charges.
(ii) Fnd the electric potential energy of the
–5 nC charge in the lower right hand corner.
b.
Shown at right are four arrangements
of charged particles, all located at the
same distance from the origin. Rank
the situations according to the net
electric potential at the origin, most
positive first. Explain your reasoning.
Let the potential be zero at infinity.
(ii) Vorigin = –2kq/d – 3kq/d = –5kq/d
c.
(i) Vorigin = 2kq/d – 9kq/d = –7kq/d
• –2q
• 2q
–9q
•
–3q
–2q
•
(i)
(iii)Vorigin = –2kq/d – kq/d – 2kq/d – 2kq/d = –7kq/d
(iv)Vorigin = –4kq/d + 2kq/d – 7kq/d + 2kq/d = –7kq/d
Then: 2q
•
•
•
• –7q
• –2q
(ii)
(iii)
(iv)
Thus: Vorigin.ii > Vorigin.i = Vorigin.iii = Vorigin.iv
Let d be the distance from any corner of the rectangle to point P.
d = (1/2)√(L2 + W2)
2q
–q
•
A rectangle with length L and width W has point charges on all four corners.
Two identical charges, located diagonally across from each other, have a
charge of q. The other two charges are also identical to each other and
have a charge –2q. Find the electric potential at point P, in the center of
the rectangle.
That is:
• –4q
• –2q
+q
•
•
–2q
L
P.
–2q
•
W
•
+q
VP.total = 2(kq/d) + 2(–2kq/d) = –2kq/d = –4kq/[√(L2 + W2)]
d. Two particles of charge +q are located on the x-axis at (5d, 0) and (7d, 0). A third particle of charge –q is
located on the x-axis at (6d, 0). What is the net electric potential caused by these three particles at the
origin?
Vtotal,origin = kq/(5d) – kq/(6d) + kq(7d) = (kq/d)[1/5 – 1/6 + 1/7] = 37kq/(210d) = 0.176kq/d
e.
Two identical point charges (q = +1.20 mC) are fixed to diagonally opposite corners of a square. Then, when
a third charge is placed at the center of the square, this causes the voltages at the empty corners of the
square to change sign without changing magnitude. Find the sign and magnitude of the charge at the center
of the square.
5
5. f. On coordinate axes, point A = (1.50, 2.00), measured in meters. Let the voltage value very far away
(“infinity”) be defined as zero (i.e. V∞ = 0). To make things simpler, just express your answers in terms of e
and k (in other words, don’t plug in numerical values for e and k).
A proton is placed at the origin:
(i) Calculate the electric field (magnitude and direction) it will cause at point A.
(ii) Calculate the magnitude and direction of the electric force it will exert on a proton located at point A.
(iii)Calculate the magnitude and direction of the electric force it will exert on an electron located at point A.
(iv)Calculate the electric potential (voltage) it will cause at point A.
(v) Calculate the electric potential energy (UE) of a proton located at point A.
(vi)Calculate the electric potential energy (UE) of an electron located at point A.
(vii) Calculate the uniform electric field (magnitude and direction) that would need to be present
(in addition to the proton), so that an electron placed at point A would have no acceleration.
6
6. a.
A charge of +4e is located at the origin. A charge of -10e is located at the point (12,0), with coordinates
given in cm. Points A and B are empty and are located along the x-axis, but neither are located between the
two charges. The electric field at point A is zero; and (using the usual convention for voltage due to point
charges: V∞ = 0), the voltage at point B is zero. Find the distance between points A and B.
b. A charge of +12e is located at the origin. A charge of -3e is located at the point (9,0), with coordinates given
in cm. Assuming the usual convention for voltage due to point charges (i.e. V∞ = 0), find a point that is
somewhere between the two points (i.e. located along the x-axis) where V = 0.
The voltage caused by any single point charge q at a
distance r is given by: V = kq/r
For qA and qB, therefore, VA = kqA/rA and VB = kqB/rB
rA
rB
We seek a point along the x-axis where VA + VB = 0.
qB = –3e
qA = +12e
X
In other words: kqA/rA + kqB/rB = 0
V=0
That is:
12ke/rA – 3ke/rB = 0 Simplifying: rA = 4rB
Thus: rA + rB = 0.09 Or:
4rB + rB = 0.09
Therefore:
rB = 0.018
The zero-voltage point is located at x = 0.0720 m
c.
q1 is an electric charge of +3e, located at the origin of a set of coordinate axes. q2 is an electric charge of –e,
located a distance d to the right of q1. The voltage value far away is zero (i.e., V∞ = 0), as is conventional.
Referring only to points located on the x-axis: What is the minimum distance you’d need to move, starting
from a point where the electric field value is zero, to arrive at a point where the voltage value is zero?
The list of known values: q1, q2, d, k, e.
I. Make a diagram of the situation.
Define the positive x-direction conventionally (to the right).
On the diagram, label 3 regions:
Region A is to the left of q1.
Region B is between q1 and q2.
Region C is to the right of q2.
A
B
d
q1
C
q2
II. Decide where E = 0 (along the x-axis) is even possible:
In region A, the direction of E1 is to the left, and the direction of E2 is to the right; they oppose,
which is necessary for their sum to be zero. However, the larger charge |q1| will be closer to any
point in this region than will the smaller charge |q2|. Since |E| = k|q|/r2, |E1| > |E2| for every point
in this region. The sum E1 + E2 could never be zero.
In region B, the direction of E1 is to the right, and the direction of E2 is also to the right; they do
not oppose, so their sum could never be zero.
7
In region C, the direction of E1 is to the right, and the direction of E2 is to the left; they oppose,
which is necessary for their sum to be zero. Morover, the larger charge |q1| will be farther from
any point in this region than will the smaller charge |q2|. Since |E| = k|q|/r2, |E1| = |E2| is possible
for some point in this region.
III.Find the x-position of the point in region C where |E1| = |E2|:
Use:
Where:
Call this xE=0.C
is given
(given)
In regions B and C, however, there are points which are farther from q1 than q2, so |V1| = |V2| is possible.
and since the charge signs are opposite, the sum V1 + V2 = kq1/r1 + kq2/r2 could be zero.
Use:
kq1/r1 + kq2/r2 = 0 k
q1 = 3e
r1 + r2 = d
d
q2 = –e
is known
(given)
Use:
Where:
kq1/r1 + kq2/r2 = 0 k
is known
q1 = 3e
(given)
r1 = d + r2
d
is given
q2 = –e
(given)
VII. Find the distance, dB, between xV=0.B and xE=0.C:
Use:
Where:
dB = xE=0.C – xV=0.B xE=0.C is from part III
xV=0.B is from part V
VIII. Find the distance, dC, between xV=0.C and xE=0.C:
Use:
Solve for: r1
Call this xV=0.B
is given
(given)
VI.Find the x-position of the point in region C where V1 + V2 = 0:
Solve for: r1
In region A, the larger charge |q1| will be closer to any point in this region than will the smaller charge |q2|.
Since |V| = k|q|/r, |V1| > |V2| for every point in this region. The sum V1 + V2 could never be zero.
Where:
is known
(given)
V. Find the x-position of the point in region B where V1 + V2 = 0:
k
|q1| = 3e
r1 = d + r2
d
|q2| = e
IV. Decide where V = 0 (along the x-axis) is even possible:
k|q1|/r12 = k|q2|/r22
dC = |xE=0.C – xV=0.C| Where: xE=0.C is from part III
xV=0.C is from part VI
IX. The minimum distance (the answer) is the lesser of dB and dC.
Solve for: r1
Call this xV=0.C
Solve for: dB
Solve for: dC
8
6. d.
q1 is a positive electric charge, located at the origin (all coordinates in m). q2 is some other electric
charge, located at the point (–2,2). The voltage value far away is zero (i.e., V∞ = 0), as is conventional.
The voltage at point (2,2) is also zero. Find the total electric field (both magnitude and direction) at the
point (–2,0). The list of known values: q1, q2, k.
(–2, 2)
(2, 2)
q2
I. Make a diagram of the situation.
X
(–2, 0)
II. If the voltage at point (2,2) is zero, this means that V1.(2,2) + V2.(2,2) = 0.
Use:
Where:
q1+
kq1/r1.(2,2) + kq2/r2.(2,2) = 0
k (known)
q1
(given)
r1.(2,2) = √(22+ 22)
r1.(2,2) = √8
Solve for: q2
This step is necessary because the magnitude of q2 was given, but not the sign.
Since q1, k and the distances r1.(2,2) and r1.(2,2) are all positive, the solution for q2
will show that it is negative.
III.The total E-field, EX, at point X is the vector sum of the fields caused at X by the two charges.
Use:
Where:
EX = E1.X + E2.X
E1.X
q1
r1.X
E2.X
q2
r2.X
Solve for: EX
= k|q1|/(r1.X)2∠180°
(Because q1 is positive, its E-field points away from it.)
(given)
= 2
= k|q2|/(r1.X)2∠90° (Because q2 is negative, its E-field points toward it.)
(from step II)
= 2
Note: The total mathematical solution for EX would involve addition of the vectors E1.X and E2.X
by x- and y- components, to get the x- and y- components of EX (call these EX.x and EX.y).
Then:
And:
|EX| = √(|EX.x|2 + |EX.y|2)
qE.X = tan-1(EX.y/ EX.x)
(the quadrant II solution)
9
6. e.
In the diagram shown, a point charge of +7.45 mC is located
at point A (the origin). Another point charge is located at
point B. The voltage at point D is zero. Find the force
(magnitude and direction) that an electron would feel if
it were placed at point C.
If the voltage at D is zero, that’s the sum of the voltages caused by the proton and other unknown
charge: The voltage at any given point in space caused by any single charge is V = kq/r, where q is
the charge (magnitude and sign) and r is the distance from the charge to the point in question.
All r values are just distances (positive) and qA is positive, so qB is negative. An electron at point C would
therefore be repelled by qB (but attracted, of course, to the proton, qA). The net force felt by an electron at C
will be the vector sum of FAC (the force on qC due to qA) and FBC (the force on qC due to qB).
Therefore, at point D: kqA/rAD + kqB/rBD = 0
So:
+y
(length units shown are meters.)
A
-x
D
3.62
-y
B
5.00
C
8.00
+x
qB = –qA(rBD/rAD)
FT.C = FAC + FBC = (k|qA||qC|/rAC2)∠180° + (k|qB||qC|/rBC2)∠0°
= (ke|qA|/82)[cos180°, sin180°] + (ke|qB|/32)[cos0°, sin0°] = [–ke|qA|/64, 0] + [ke|qB|/9, 0]
9
-19
-6
= [keqA(1.38/32.58 – 1/64), 0] = [(8.99 x 10 )(1.60 x 10 )(7.45 x 10 )(1.38/32.58 – 1/64), 0]
f.
-16
-16
= [2.86 x 10 , 0] = 2.86 x 10
N∠0°
+y
In the diagram shown, a negative point charge,
qA = -21.0 µC, is located at point A (the origin).
Another point charge, qB, is located at point B.
The x-component of the electric field at point C
is -543 N/C. What would be the electric potential
energy of an electron placed at point D?
-x
A
D
3.62
-y
B
5.00
C
8.00
+x
The E-field at C is the sum of the fields caused by qA and qB: EC = EqA.C + EqB.C = 543 N/C∠180°
The field magnitude at any given point in space caused by any single charge q is E = k|q|/r2, where |q| is the charge magnitude, and r is the distance from the charge to the point in question.
Since point C is to the right of A, and qA is negative, the direction of EqA.C will be to the left (negative).
Calculate that field strength by itself: = (8.99 x 109)(21.0 x 10-6)/(82) = 2950 N/C∠180°
So the proper summation of the field at point C is: 2950 N/C∠180° + k|qB|/rBC2 ∠0° = 543 N/C∠180°
Since the net field is only 543 N/C∠180°, EqB.C must oppose EqA.C. Therefore qB must be a positive charge.
Note that all y-components for all these vectors are zero; only the x-components are relevant.
Thus: 2950cos180° + k|qB|/rBC2cos0° = 543cos180°
Or:
qB =
–2950 + kqB/rBC2 = –543
(rBC2)2407/k
-6
= 2.410 x 10 C
Solve this for qB....
Now, just calculate the voltage at point D due to these two charges:
VD = VqA.D + VqB.D = kqA/rAD + kqB/rBD
And the UE of an electron then placed at D would be given by:
UE = q0VD
= (e-)VD = (e-)[kqA/rAD + kqB/rBD]
-19
9
–6
= –(1.60 x 10 ){(8.99 x 10 )(–21.0 x 10 )/3.62
= 5.83 x 10
9
+ [(2.410 x 10-6)(8.99 x 10 )/(1.38)]}
–15
J.
10
7. a. A positive point charge is located at the origin. Assigning zero voltage to a very distant point (i.e. let
V∞ = 0), the resulting voltage at the point (3,0) has a value V1. Write an expression for the voltage at (0,2).
V1 = kq/r1 = kq/3
Therefore: V(0,2) = kq/2 = k(3V1/k)/2 = (3/2)V1
So:
q = 3V1/k
b. A positive point charge is located at the origin. Letting V∞ = 0, the voltage at the point (2,0) has a known
value V. Write an expression for the electric field (magnitude and direction) at the point (0,–3).
V = kq/r = kq/2
2
So:
Therefore: E(0,-3) = kq/3 = kq = k(2V/k)/9 = (2/9)V
Thus: E(0,-3) = (2/9)V∠270°
q = 2V/k
And the direction of E(0,-3) is away from the positive charge, q, at the origin.
c. A uniform electric field of known magnitude E is directed along the –x-axis. The voltage at the origin is a
known value V. Write an expression for the voltage value at the point (3,0), assuming all coordinates are in
meters.
The direction of E is ∠180°. This is the direction of decreasing voltage.
The value of E is given by E = –DV/Ds||, where Ds|| = scosf , and f = ∠Es.
Thus, Ds|| = scosf = –3 m.
Moving from (0,0) to (3,0), s is 3 m and in the opposite direction as E, so f = 180°, and cosf = –1.
And so, moving from (0,0) to (3,0):
Therefore, moving from (0,0) to (3,0): E = –DV/Ds||
E = –DV/(–3)
DV = 3E
DV = V(3,0) – V(0,0) = V(3,0) – V = 3E
V(3,0) = V + 3E
d. The sketch shows two charged conductors. The other (empty) points have the following potential values:
A: 200V B: 0V
C: –50V
D: –200V
D•
E: –200V F: –200V
G: –400V
H: –400V
(i)
Sketch a set of possible equipotential paths.
Path nearest to L should contain point A. The next
path should contain B; then the next path, C.
The next path should contain D, E and F.
The path nearest to R should contain G and H.
C•
A•
B•
F•
G•
R
H•
L
(ii) Which conductor is positively charged? L
(iv)At which point would an electron have the greatest potential energy? G or H
E•
(iii)What is the magnitude of the potential difference between points A and H? 600 V
(v) What is the direction of the electric field at point E? Approximately toward the center of R.
(vi)How much work does the electric field do on a +6 μC point charge that moves from B to F to D to A?
Wfield.BA = –DUE.BA = –qDVBA = –(6 x 10-6)(200 – 0) = –0.00120 J
e. Shown here are three equipotential surfaces.
All cover the same size region of space.
(i) Rank the arrangements according to the magni-
tude of the electric field present in the region.
(ii) Label the direction of each electric field.
(i) |E1| > |E2| = |E3|
(ii) E1: up the page
E2: up the page
(1)
(2)
(3)
————— 20 V ————— –140 V ————— –10 V
————— 40
————— 60 ————— –120 ————— –30
————— 80
————— 100 ————— –100 ————— –50
E3: down the page
11
y
A (1.00, 5.00)
B (7.00, 3.00)
x
(i) The total E-field is a (vector) sum here of two uniform fields: ET = E1 + E2 E2 is given: 300∠270°
E1 = E1∠0°
E1 = –DV/(Dscosf),
where DV is the voltage difference between the plates,
and Ds = 0.10 m, and f = 0° (proceeding from left to right).
Thus: E1 = –(–40)/[(0.10)(cos0°)] = 400 V/m
So: E1 = 400∠0°
Thus: ET = 400∠0° + 300∠270° = [400, 0] + [0, –300] = [400, –300] = 500 N/C ∠–36.87°
Any equipotential path is at right angles (±90°) from the total E-field.
Thus, possible equipotential paths would include ∠–127° (= ∠233°) and ∠53.1° (= ∠–307°).
(ii) The force on the electron: F = qE = (–e)500∠–36.87° = 500e∠143° = 8.00 x 10-17 N ∠143°
(iii)Moving from A to B, the electron’s displacement vector, Ds, would be given by
Ds = √[(.07–.01)2 + (.03–.05)2]∠tan-1[(.03–.05)/(.07–.01)] = 0.06325∠–18.43°
Therefore, f = ∠Es = –18.43° – (–36.87°) = 18.44°
The work necessary to accomplish the displacement, Ds, is the change in potential energy, DUE.
And DUE is given by:
DUE = qDV = q(–EDs·cosf) = (–e)[–500(0.06325)cos(18.44°)]
Therefore: W = (–1.60 x 10-19)[–500(0.06325)cos(18.44°)] = 4.80 x 10-18 J
b. In the coordinate system shown, a proton is located at the origin,
and an electron is located at the point (d,d). Point X is an empty
(0, d) X
point, located at (0,d). Let V∞ = 0.
(i) Write an expression for the total voltage at point X.
(ii) Briefly re-sketch the situation, and then on your sketch,
draw an equipotential path that contains point X.
(iii)Write an expression for the total electric field (both magnitude
and direction) at point X.
e+
(iv)Write an expression for the total electric force (both magnitude
and direction) that would be exerted on an electron if it were placed at point X.
(iii)Eproton.X = ke/d2∠90°:
Add the two vectors:
(0, d) X
(d, d)
e-
Eelectron.X = ke/d2∠0°: ETotal.X = Eproton.X + Eelectron.X
e-
(i) VTotal.X = Vproton.X + Velectron.X = k(e)/d + k(-e)/d = 0
(ii) An equipotential path is shown here. (Any point equidistant from the two
charges would have a 0 voltage value.)
(d, d)
= ke/d2∠90° + ke/d2∠0°
e+
(d, 0)
= [0, ke/d2] + [ke/d2, 0] = [ke/d2, ke/d2] = (ke/d2)(√2)∠45°
(iv) Felectron.X = (-e)ETotal.X = (-e)(ke/d2)(√2)∠45° = (ke2/d2)(√2)∠225°
12
| | | | | | | | | | | | | | | | | | |
++++++++++++++++
8. a. Initially, there is a set of parallel plates, as shown here, separated
by a distance of 10.0 cm, and the voltage difference between the
plates is 40.0 V. The electric field between the plates is uniform.
Then a second uniform field, E2 = 300 N/C ∠270°, is also applied
to the entire region shown (and you may assume that the charges
and positions of the plates remain unchanged even with the presence
of E2). Points A and B are empty. All coordinates are given in cm.
(i) Give one direction (an angle, measured from ∠0°) that you
could move (in the x-y plane) from point A along an equipotential
path.
(ii) What force (magnitude and direction) would be exerted on an
electron placed at point B?
(iii)What work would be required to move an electron from A to B?
The “slope” of the contours of the potential (voltage) is the strength (magnitude) of the E-field. In other words,
the ratio of the change in the voltage to the distance you travel along the E-field lines (in the direction of the Efield) gives the magnitude of the E-field: E = –DV/Ds||. (The negative sign shows that DV is negative when E
and Ds|| are positive.) Of course, this means that if you move perpendicularly to the E-field lines, the voltage is
constant; that is an equipotential path. If E = –DV/Ds||, then DV = –EDs||. We know that E = 6300 N/C (in the +xdirection). The question is, what is Ds||? What distance in each case, are we moving along an E-field line?
That is, what is EDs·cosf, where f is the angle between E and Ds?
y
A (0, 0)
C (2.50, 4.00)
B (2.50, 0)
x
| | | | | | | | | | | | |
A uniform electric field of magnitude 6300 N/C is
produced between two parallel plates of charge (i.e.
a parallel-plate capacitor), as shown. For the points
given (all units in cm), find these voltage differences:
(i) VB – VA
(ii) VC – VB
(iii)VA – VC
+++++++++++++
8. c.
(i) Moving from A to B, we move 2.50 cm along an E-field line in the direction of the E-field.
So Ds|| = +2.50 cm (= 0.025 m). And so VB – VA = DV = –EDs|| = –(6300)(0.025) = –158 V.
(ii) Moving from B to C, we do not move at all along an E-field line in the direction of the E-field
(not even backwards). We are moving perpendicularly to the field. So Ds|| = 0.
And so VC – VB = DV = –EDs|| = –(6300)(0) = 0 V.
(iii)Moving from C to A, although we move at a diagonal through the E-field, only the component of
the motion along the direction of the E-field would require work to be done on a charge. So only
the x-portion of the motion changes the voltage. Thus Ds|| = –2.50 cm, since it’s in the direction
opposite to the direction of the E-field. So VA – VC = DV = –EDs|| = –(6300)(–0.025) = 158 V.
d. (i) Consider a set of 3-D coordinate axes (x-y-z). At x = 1.00 m, the voltage is 4.00 V. At x = 3.00 m, the
voltage is also 4.00 V. Assuming that there is a steady, uniform non-zero electric field over this entire
region, give four possible directions of that electric field.
Any two points in a uniform electric field that have the same voltage are on the same equipotential path. Clearly, the x-axis is an equipotential path (4.00 V). So any direction perpendicular to the x-axis is a possible
direction for the E-field. There are an infinite number of such directions—here are four easy ones
Positive y-direction; negative y-direction; positive z-direction; negative z-direction.
E = –DV/Ds|| (Ds|| = scosf , and f = ∠Es.) Moving from (0,0) to (0,–3), s is 0.03 m and f = 150°.
Thus, Ds|| = scosf = (0.03)cos150°. And so, moving from (0,0) to (0,-3):
E = –DV/Ds||
Or: 256 = –DV/[(0.03)cos150°]
Or: DV = –256(0.03)cos150° = (+)6.6511 V
And then: W = DUE = qDV = (e)(6.65) = (+)1.06 x 10-18 J
(ii) If there is a steady, uniform electric field (in the x-y plane) given by E = 256 N/C ∠60°, how much work
would be required to move a proton from the origin to the point (0,–3)? (Coordinates are in cm.)
e. At (1, 0), the voltage is 4.00 V. At (3, 0), the voltage is 0.00 V. (Coordinates given are in cm.) Assuming
that, all over this region of the x-y plane, there is a steady, uniform electric field parallel to the x-axis:
(i) What is the magnitude and direction of that electric field?
(ii) How much work is required to move an electron from the origin to (6, 0)?
The direction of E is the positive x-direction (∠0°), because that is the direction of decreasing voltage.
The magnitude of E is given by E = –DV/Ds||, where Ds|| = scosf , and f = ∠Es.
Moving from (1,0) to (3,0), s is 0.02 m and in the same direction as E, so f = 0°, and cosf = 1.
Thus, Ds|| = scosf = s = 0.02 m.
And so, moving from (1,0) to (3,0): E = –DV/Ds|| = –(–4.00 V)/0.02 m = 200 V/m. E = 200 V/m ∠0°
Moving from (0,0) to (6,0), s is 0.06 m and in the same direction as E, so f = 0°, and cosf = 1.
Thus, Ds|| = scosf = s = 0.06 m. And so, moving from (0,0) to (6,0):
E = –DV/Ds||
200 = –DV/0.06
DV = –200(0.06) = –12.0 V
And then: W = DUE = qDV = (–e)(–12.0) = (+)1.92 x 10-18 J
13
8. f. Throughout the region
shown, there is a steady, uniform E-field.
y
B (–1, 1)
•
The electric potential (“voltage”) at the origin
is 500 V. The voltage at point A is 300 V.
The voltage at point B is 500 V. All coordinates are in m.
A (2, 1)
•
(i) T/F/N? The voltage at point C
is 300 V. Justify your answer fully.
x
30
0
V
50
C (5, –2)
•
0
V
(ii) How much work must you do to move an electron from the origin to point A?
(iii)Find the acceleration (magnitude and direction) of a proton released from rest at the origin.
14
8. g.
There is a proton located at the origin and an electron located at the point (–2, –2). (Coordinates are in
meters.)
(i) Give one direction (angle) you could walk from the point (0,–2) along an equipotential path.
(ii) Find the electric field strength at (–2, 2).
(iii)What electric potential energy change would the proton undergo by moving from the origin to (–1, 1)?
(i)
The two charges values are equal in magnitude but opposite in sign,
any point equidistant from both, such as (0, –2), will have zero voltage.
So two example equipotential paths from (0, –2) would be ∠135° and
∠–45°, as shown by the dotted line.
And:
• (–1, 1)
(ii) The net E-field at (–2,2) is the vector sum of the E-fields caused by the
two charges: E = Ep + Ee So:
• (–2, 2)
E = (ke/8)[cos135°, sin135°] + (ke/16)[cos270°, sin270°]
= (ke/16)[–1.414, 0.414]
2
2 1/2
-1
= (ke/16)[(–1.414) + (0.414) ] ∠tan [0.414/(–1.414)]
= 1.32 x 10-10 N/C∠tan-1[0.414/(–1.414)]
(iii)DUE is given by:
DUE = qDV
So:
DUE = qDV = e[k(–e)/(√10) – k(–e)/(√8)]
•
Ep = ke/8∠135° and Ee = ke/16∠270°
The proton relocating:
Vi = k(–e)/(√8) and Vf = k(–e)/(√10) • (–2, 2)
• (0, –2)
= (ke2)[1/√8 – 1/√10] = 8.59 x 10-30 J
h. There is a uniform electric field directed at 150° above the positive x-axis (and the field is in the x-y plane).
The voltage value measured at the origin is –6.00 V. The voltage value measured at (3,–4) is 9.00 V.
(Coordinates are given in cm.)
(i) Give one direction (angle) you could walk from the origin along an equipotential path.
(ii) Find the force (magnitude and direction angle) exerted on an electron in this field.
(iii)What minimum distance would an electron need to be moved from the origin to gain 3.00 x 10-18 J in
electric potential energy?
15
9. Evaluate (T/F/N) each statement. Justify your answers fully with a valid mix of words, drawings and calculations.
The graph shown here applies to statements a and b, below.
You may assume that V does not vary in the y- or z- direction.
a. A proton located at x = –1.00 mm would experience an electric
force of 1.07 x 10–12 N, in the +x direction.
False. Fx = qEx = q(–dV/dx) = q(–DV/Dx)
= –(1.60 x 10-19)[(20,000)/(.003)]
= –1.07 x 10-12 N
This is in the –x direction.
V (kV)
20
10
–4
–2
2
4
x (mm)
b. It would require a positive (non-zero) amount of external work to move an electron (at constant speed)
from x = 1.00 mm to x = 4.00 mm.
True. Wext = q(DV) = (–e)(Vf – Vi)
= –(1.60 x 10-19)[0 – 20,000] = 3.20 x 10-15 J This is positive work.
c. You are standing at the origin in a region where a uniform electric field exists: E = [1.50, 2.50, –8.00] N/C.
If you then move directly toward the point [-6.00, 10, 2.00], you will be moving along an equipotential path.
True. Your path from the origin to [–6, 10, 2] is perpendicular to the field (thus an equipotential),
which can be demonstrated via E • Ds = |E||Ds|·cosq = 0 (thus q = ± 90°):
E • Ds = [1.50, 2.50, –8.00] • [–6.00, 10, 2.00] = (1.5)(–6) + (2.5)(10) + (–8)(2) = 0
The following situation applies to the next two statements (d and e).
Object 1 is a conductive sphere of radius 1.00 cm, centered at (–4.00, 0) and charged to a voltage of +10,000 V.
Object 2 is point charge of –5.00 nC, located at (3.00, 0). Both objects are fixed (stationary).
All coordinates are given in cm, and V∞ is defined as 0. d. If you were to place an additional point charge of –2.23 nC at (0, –2.50), this would result in a total voltage
value of 0.00 at the origin.
False. V1(r) is given by V1.R(R/r) = (10,000)(.01/.04) = +2,500 V
V2(r) is given by kq2/r = (8.99 x 109)(–5.00 x 10-9)/.03 = –1498.3 V
V3(r) is given by kq/r = (8.99 x 109)(–2.23 x 10-9)/.025 = –802.0 V
2500 + (–1498) + (–802) = 200.3 (not 0)
e. Instead of part d [i.e. no additional charge placed at (0, –2.50)], if you place a proton at the origin and
release it from rest, it will be moving at about 4.89 x 105 m/s as it passes through the point (1.00, 0).
True. DKT + DUE = 0
Or: (1/2)mpvf2 + (eVf ) – (eVi ) = 0
Thus: vf = √[(2e/mp)(Vi – Vf)]
Vi = 2500 – 1498 = 1001.7 V (from above) 9
-9
Vf = (10,000)(.01/.05) – (8.99 x 10 )(5.00 x 10 )/.02 = –247.5 V
So: vf = √{[(2(1.60 x 10-19)/(1.67 x 10-27)](1001.7 – –247.5)} = 4.89 x 105 m/s
16
9. f.
Positive point charges q1 and q2
are fixed at the positions shown along the x-axis.
A third positive charge, q3 (mass = m3), is released
from rest at the origin, and it then moves freely
(moving along the x-axis only), influenced only
by the other two charges.
(ii) What is q3’s greatest speed?
(iii)At what x-position does q3 most closely approach q2?
(–d1, 0)
(d2, 0)
q1
q2
x
(i) At what x-position does q3 attain its greatest speed?
Note:
y
This problem contains no numeric data—symbols only. But rather than actually solving algebraically
for the required values, you may, if you wish, simply give the necessary equations and describe step-bystep what/how to solve (much like the Equations and Solve steps of the ODAVEST method). But you
must be specific for every step. A person who knows only math (no physics) should be able to calculate
correct answers using your instructions.
You may consider the following known values (they may appear in your solutions): q1, d1, q2, d2, q3, m3, k, e0
(i)
The charge q3 will accelerate one way or the other due to a net E-field (i.e. the combined effects of the
two other charges’ E-fields) until it reaches the point where the net E-field is zero. Past that point, the
net force would be opposite the direction of motion, so that point is where q3 begins to slow down;
it’s the point when it has its greatest speed.
So we simply need to find the x-location where Etotal = E1 + E2 = 0
That is, we want the fields E1 and E2 to be equal in magnitude;
So we would solve this equation for xv.max:
kq1/(d1 + xv.max)2 = kq2/(d2 – xv.max)2
There will be two mathematical solutions, but only –d1 < xv.max < d2 is physically valid;
q3 would never actually reach either charge, because that charge’s force would approach ∞.
(ii) When q3 is first released, v3 = 0; all its mechanical energy is in the form of UE:
Emech.i = UE.i = kq1q3/d1 + kq2q3/d2
At the point of maximum speed (xv.max from part a, above), Emech is partly UE and partly KT:
Emech.f = UE.f + KT.f = kq1q3/(d1 + xv.max) + kq2q3/(d2 – xv.max) + (1/2)m3v3.max2
There is no external work being done here (Wext = 0), so Emech.f = Emech.i
So (using xv.max as a known value now) we would solve this equation for v3.max:
kq1q3/(d1 + xv.max) + kq2q3/(d2 – xv.max) + (1/2)m3v3.max2 = kq1q3/d1 + kq2q3/d2
(iii)When q3 is first released, v3 = 0; all its mechanical energy is in the form of UE:
Emech.i = UE.i = kq1q3/d1 + kq2q3/d2
When q3 reaches its closest approach to q2, v3 = 0, also; again, all its mechanical energy is in UE:
Emech.f = UE.f = kq1q3/(d1 + xmax) + kq2q3/(d2 – xmax)
There is no external work being done here (Wext = 0), so Emech.f = Emech.i
So we would solve this equation for xmax:
kq1q3/(d1 + xmax) + kq2q3/(d2 – xmax) = kq1q3/d1 + kq2q3/d2
There will be two solutions: xmax = 0 and 0 < xmax < d2 . Both are physically valid:
xmax > 0 if E1 > E2 at the origin. xmax = 0 if E2 > E1 at the origin.
17
10. a. An infinite nonconducting sheet has a surface charge density h = 0.10 μC/m2 on one side.
How far apart are equipotential surfaces whose potentials differ by 50 V?
The field near an infinite sheet of charge is uniform, with a strength given by E = h/(2e0).
Therefore: d = 2e0DV/h = 2(8.85 x 10-12)(50)/(0.10 x 10-6) = 0.00885 m For a uniform field, the distance between equipotentials is given by d = DV/E
(= 8.85 mm)
b. Textbook problem 26.53, page 740.
The potential difference across the fully charged capacitor is that of the battery: DVC = 9.00 V.
And then Q = (e0A/d)DVC = [(8.85 x 10-12)(.022)/(.001)](9.00) = 3.19 x 10-11 C
For the wider separation, Q has not changed, and we know that DVC = Qd/(e0A)
= (3.19 x 10-11)(.002)/[(8.85 x 10-12)(.022)] = 18.0 V
c. A parallel plate capacitor with plates of area A and plate separation d is charged so that the potential
difference between its plates is V. If the capacitor is then isolated and its plate separation is decreased to
d/2, by what factor is the potential difference between the plates changed?
DVC = Qd/(e0A), so if we multiply d by 1/2 we also multiply DVC by 1/2.
d. The potential in the xy-plane is given as (x2 + y2)-1.
What is the direction of the electric field at the point (d, 0), where d > 0?
Exy = [–∂(V)/∂x, –∂(V)/∂y] = [–2x/(x2 + y2)2, –2y/(x2 + y2)2]
e.
A uniformly charged rod (total net charge Q) lies along the x-axis as
shown.
Defining V(∞) = 0, show how you would find the electric potential at point P,
which is a distance h above the x-axis. You do not need to perform the integration
—just “build the integral.”
qE = tan-1[Ey/Ex] = tan-1[–y/x] = 0 at the point (d, 0).
Sum up all the differential contributions of potential by each
differential of charge along the rod:
y
P
h
dV = kdq/r, where: dq = ldx = (Q/L)dx and: r = √(x2 + h2)
Thus: The integral:
2
x
2
dV = k[(Q/L)[dx/√(x
+ h )]
L/4
V = (kQ/L)
∫
–3L/4
L
[1/√(x2 + h2)]dx
L/4
The above is sufficient to answer the question. In case you’re curious, here’s the rest of the math.
∫
L/4
V = (kQ/L) –3L/4 [1/√(x2 + h2)]dx
2
2
L/4 = (kQ/L)ln[x + √(x + h )]–3L/4
= (kQ/L){ln[L/4 + √(L2/16 + h2)] – ln[–3L/4 + √(9L2/16 + h2)]}
= (kQ/L)ln{[L/4 + √(L2/16 + h2)]/[–3L/4 + √(9L2/16 + h2)]}
= (kQ/L)ln{[L + √(L2 + 16h2)]/[–3L + √(9L2 +16h2)]}
= (kQ/L)ln({1 + √[1 + (4h/L)2]}/{–3 + √[9 +(4h/L)2]})
18
10. f. Consider a long, positively charged wire of radius R. Applying Gauss’s Law (ex. 24.5, pages 674-675) has
shown that the magnitude of the electric field at a distance r (> R) from the center of the wire is given by
E = λ/(2pe0r).
(i) What is the electric potential for r > R? (Use the zero of electric potential at r = R.)
(ii) Now, assume a proton is fired from some distance ri towards the wire along a trajectory perpendicular
to the wire. What initial speed must the proton (mass = mp) have if it is to momentarily stop, then
reverse its direction of motion, at a distance of 3R from the center of the wire?
(iii)How would the answer to part (ii) change if the proton is fired along a trajectory that is not
perpendicular to the wire? Explain.
(i) V(r) = –
∫
ˆr
R
E(r)dr = –λ/(2pe0)
ˆr
R
(1/r)dr = –λ/(2pe0)[ln(r/R)]
(ii) For the proton:
UE.i + KT.i = UE.f + KT.f
∫
(1/2)mpvi2
That is: KT.i = UE.f – UE.i
= –keλ/(2pe0)[ln(R/R) – ln(3R/R)] = keλln(3)/(2pe0)
Thus: vi = √[keλln(3)/(pe0mp)]
(iii)The proton would not completely stop; only its velocity in the r direction is affected by the wire’s field—
with the work being done as above. So the components of initial velocity in the q or z directions (at least one
of which is non-zero in this scenario) would require a total initial speed greater than when the total velocity
was along the r-axis only.
g.
The figure shows a graph of V versus x in a region of space.
The potential is independent of y and z.
(i) Find the electric field at (0.2, 0, 0.4); coordinates in m.
(ii) Sketch a graph of Ex versus x.
19
11. a.
A 4 μF capacitor is charged until it acquires a potential difference of 500 V across its plates, then the
voltage source is removed.
(i) What is the charge on the capacitor? Q = CDV = (4 x 10-6)500 = 2.00 x 10-3 C
(ii) What is the energy stored on the capacitor? UE = (1/2)CDV2 = (1/2)(4 x 10-6)5002 = 0.500 J
c.
A 4 μF and a 14 μF capacitor are connected in series, and the series arrangement is connected in
parallel to a 27 μF capacitor. How much capacitance would a single capacitor need to replace the
three capacitors?
Combine the two series capacitors: 1/Cbranch = 1/4 + 1/14. Or: Cbranch = (4)(14)/(14 + 4) = 3.11 μF
Combine the two parallel branches: Ctotal = C27 + Cbranch = 27 + 3.11 = 30.1 μF
b.
d.
N identical capacitors, where N > 3, are connected in parallel, and the combination is connected
across the terminals of an ideal battery (so disregard any resistance). In this configuration, let the
charge on each capacitor be called Q. If one capacitor were then removed, what would be the charge
on each capacitor?
The charge on each remaining identical capacitor would still be Q, because the battery would still be
providing the same voltage difference to each.
Textbook problem 26.57, page 740.
Add the two parallel capacitors:
C12 = 4 + 12 = 16 μF
Combine that result in series with C3: 1/Ctotal = 1/2 + 1/16. So: Ctotal = (2)(16)/(16 + 2) = 1.78 μF
But since C12 and C3 are in series, they each have the same charge (the total charge for the circuit).
Therefore: Q12 = Q3 = Qtotal = CtotalDVtotal = (1.78 x 10-6)(9) = 16.0 µC
So: DV3 = Q3/C3 = (16.0 x 10-6)/(2.00 x 10-6) = 8.00 V
Therefore: DV1 = DV2 = 1.00 V
So: Q1 = C1DV1 = (4.00 x 10-6)(1.00) = 4.00 µC
And: Q2 = C2DV2 = (12.0 x 10-6)(1.00) = 12.0 µC
e. What would be the voltage across a 10.0-F capacitor if it stored the same amount of energy as
a 1000-kg car would have when moving at 40.0 m/s?
(1/2)CDV2 = (1/2)mv2 Or: DV = √[mv2/C] = √[(1000)(40)2/(10)] = 400 V
f. Textbook problem 26.64, page 740.
Find the difference in energy stored:
UE.i = Qi2/(2Ci) = (4.00 x 10-3)2/[2(5.00 x 10-6)] = 1.60 J
UE.f = Qf2/(2Cf) = (4.00 x 10-3)2/[2(2.00 x 10-6)] = 4.00 J
Thus: Wext = DUE = UE.f – UE.i = 4.00 – 1.60 = 2.40 J
g.
A parallel-plate capacitor has a capacitance of 2.50 nF. It is connected to the terminals of a battery
which has a very small emf. After charging the capacitor, there is only one excess electron on the
negatively-charged plate of the capacitor. The capacitor is disconnected from the battery and the
insulator is removed from the region between the plates so that the one excess electron can move to the
positively-charged plate. How fast is the electron moving when it strikes the positively-charged plate?
DUE + DKT = 0
Or:
Q2/(2C) = (1/2)mv2
Therefore: v = √[Q2/(mC)] = √{(1.60 x 10-19)2/[(9.11 x 10-31)(2.50 x 10-9)]} = 3.35 m/s
h. Evaluate (T/F/N) the following statement. Justify your answer fully.
If you widen the separation distance between the plates of a charged parallel-plate capacitor (without
altering the charge on the plates), you are increasing the total electric potential energy stored, and you are
decreasing the capacitance value of the capacitor, but you are not changing the energy density in the space
between the plates.
True. C = Q/DV = Q/(Ed) = Q/[(h/e0)·d] = Ae0/d So: Increasing d decreases C.
UE = (1/2)CDV2 = Q2/(2C). But C decreases with increasing d.
So: UE increases.
2
2
2
2
uE = UE/VolC = [Q /(2C)]/(Ad) = Q /[(2(Ae0/d)Ad)] = Q /(2A e0) So: uE does not change.
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11. i. Consider this circuit of fully-charged, identical 3.00-mF capacitors.
(i) What would be the magnitude of the reading on an accurate
and properly operated voltmeter whose probes were touched
to points A and B?
(ii) By what maximum factor could you increase the energy
storage by rearranging these capacitors (then connecting the
modified circuit to the same battery and allowing full charging)?
r
600 V
A
B
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12. a. Which of the following situations results in a conventional electric current that flows westward?
(i) a beam of protons moves eastward
(ii) an electric dipole moves westward
(iii)a beam of electrons moves westward
(iv)a beam of electrons moves eastward
(v) a beam of neutral atoms moves westward
Conventional electric current, I, is defined as the direction of positive charge flow (which is always opposite
to the flow of alectrons). Therefore, only (iv) would result in an eastward current, I.
b. How many electrons flow from a battery that delivers a current of 3.0 A for 12 s?
(3.0 C/s)(12 s)/(1.60 x 10-19 C/electron) = 2.25 x 1020 electrons
c. A 1.5 m length of gold wire is connected to a 1.5 V battery, and a current of 4.2 mA flows through it.
What is the radius of the wire? The resistivity of gold is 2.24×10-8 Ω·m.
DV/I = R = rL/A = rL/(pr2)
r = √[rLI/(pDV)] = √{(2.24 x 10-8)(1.5)(.0042)/[1.5(p)]} = 5.47 x 10-6 m (= 5.46 µm)
d. Explain briefly what you can predict about the temperature dependence of resistivity based on the
microscopic model of current, and why resistivity is dependent on the type of metal (also based on the
microscopic model).
As per our discussion in Class 15 (and see also the full summary in After class 15), the resistivity, r, in our
microscopic model of current is given by
r = [mmol/(rmasst)][me/(Ne2)], where mmol is the molar mass of the metal and rmass is its mass density.
Those two factors determine the density/availability of the conduction electrons. And t is the mean time between
collisions between the conducting electrons and the ionic lattice through which they are drifting in response to the
prevalent E-field in the conductor. As temperature rises, the value of t decreases (more thermal activity), so the
resistivity of the metal increases (note that t is in the denominator of the expression for r).
e. Determine the length of copper wire that has a resistance of 0.172 Ω and a cross-sectional area of
1 x 10-4 m2. The resistivity of copper is 1.72 x 10-8 Ω·m.
L = AR/r = (1.00 x 10-4)(0.172)/(1.72 x 10-8) = 1000 m
R = rL/A I = eie = e[N/mmol]rmassA·vd
f. The 12-gauge copper wire in a typical residential building has a cross-sectional area of 3.31 x 10-6 m2.
If it carries a current of 10.0 A, what is the drift speed of the electrons? Assume that each copper atom
contributes one free electron to the current. The density of copper is 8.95 g/cm3 and the molar mass is
63.5 g/mol.
Therefore: vd = I·mmol/(eNrmassA) = (10.0)(.0635)/[(1.60 x 10-19)(6.02 x 1023)(8950)(3.31 x 10-6)]
g.
= 2.23 x 10-4 m/s
Evaluate (T/F/N) the following statement. Justify your answer fully with any valid mix of words, drawings and
calculations.
A capacitor is discharged by connecting its plates via a 25.0-cm wire. The discharge completes in 4.00 ns.
Therefore, average drift speed of the electrons during the discharge was about 6.25 x 107 m/s.
False. Typical current electron drift speeds are around 10-4 m/s. The electrons that reach the positive plate of the
capacitor are not the same ones that leave the negative plate. After all, the entire wire is full of conductive
electrons, which all shift—just as flowing water shifts in a full pipe (and the water molecules entering the pipe are
not the same molecules that are exiting from the other end).
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