HMWK 1

HMWK 1
Ch. 20 Q 13; P 1, 15, 24, 32, 49, 51, 57, 62, 65
Ch. 21 Q 17; P 8, 27, 38, 46, 48, 54, 56, 62, 73
Q20.13. Reason: (a) E1 < E2 , because the contributions from the two charges cancel at point 1 (E1 = 0), but
the field contribution from each charge points to the right at point 2.
(b) E1 > E2 , because the contributions from each charge point in the same direction (to the right) at point 1 but
point in opposite directions and somewhat cancel at point 2 (the cancellation is not complete since point 2 is
closer to the negative charge than the positive one).
(c) E1 = E2 . The contributions from the different charges are in opposite directions at each of the two points, but
in each case the nearest charge to the point makes the stronger field to the left. The contribution from the other
charge reduces the field a bit, but there is still a net field to the left in each case. The magnitudes of the field
strength
are equal.
(d) E1 < E2 , because the contributions from the two charges cancel at point 1 (E1 = 0), but at point 2 both field
contributions have an upward component that doesn’t cancel.
(e) E1 < E2 . At point 1 the contributions are in opposite directions and partially cancel. While point 2 is farther
from the charge on the left, both contributions are in the same direction, so the field is stronger there.
(f ) E1 > E2 . At point 1 the contributions are in the same direction (to the right), whereas at point 2 they partially
cancel because they are in opposite directions (not to mention that point 2 is farther away from the positive
charge).
Assess: It is worth spending a few minutes to get comfortable with all these cases.
There are various physics software packages that allow you to map the fields around
various charge distributions; they would be good to play with also.
P20.1. Prepare: We will use the charge model. An electron has a negative charge of magnitude
1.6 × 10 −19 C.
Solve: (a) In the process of charging by rubbing, electrons are removed from one material and transferred to
the other because they are relatively free to move. Protons, on the other hand, are tightly bound in nuclei. So,
electrons have been removed from the glass rod to make it positively charged.
(b) The number of electrons removed is
5 × 10−9 C
10
= 3.1 × 10
1.6 × 10 −19 C
Assess: A large number of electrons are needed to create a modest charge.
P20.15. Prepare: Please refer to Figure P20.15. The charged particles are point charges. The charge q2 is in
static equilibrium, so the net force on q2 is zero. If q2 is positive, q1 will have to be positive to make the net force
zero on q2. And, if q2 is negative, q1 will still have to be positive for q2 to be in equilibrium. We will assume that
the charge q2 is positive. For this situation, the force on q2 by –2 nC charge is to the left and by q1 is to the right.
Solve: We have
r
r
Fn e t o n q 2 = Fq1 o n q 2 + F−2 n C o n q 2 =
1
|q1 ||q 2 |
4 πε0 ( 0.2 m)
+
2 , + x-direc tion
(2 × 10
1
4 πε 0
−9
C )| q 2 |
(0.10 m )
2
,− x -direction = 0 N/C
Thus,
−9
q1
2 × 10 C
2 −
2 = 0 N/C
(0.10 m)
(0 .2 m)
q1 = 8 nC
Assess: If the charge q2 is assumed negative, the force on q2 by –2 nC charge is to the
right and by q1 is to the left. The magnitude of q1 remains unchanged.
P20.24. Prepare: The electric field is that of the two charges placed on the y-axis. Please refer to Figure
P20.24. We denote the upper charge by q1 and the lower charge by q2. Because both the charges are positive,
their electric fields at P are directed away from the charges.
Solve: The electric field strength of q1 is
E1 = K
|q1| (9.0 ×10 9 N m2/C2 )(1 ×10 −9 C)
=
= 1800 N/C
2
2
r21
(0.050 m) + (0.050 m)
Similarly, the electric field strength of q2 is
E2 = K
| q2| (9.0 × 10 9 N m2/C2 )(1 ×10 −9 C)
=
= 1800 N/C
2
2
r 22
(0.050 m) + (0.050 m)
We will now calculate the components of these electric fields. The electric field due to q1 is away from q1 in the
fourth quadrant and that due to q2 is away from q2 in the first quadrant. Their components are
E1 x = E1 cos 45°
E1 y = −E 1 sin 45 °
E2 x = E 2 cos 45 °
E2 y = E 2 sin 45°
The x and y components of the net electric field are:
( E ne t )x = E 1 x + E2 x = E1 cos 45° + E 2 cos 45° = 250 0 N/C
( E n et ) y = E 1 y + E 2 y = −E 1 sin 45° + E 2 sin 45° = 0 N/C
E n et a t d o t = ( 2500 N/C, along + x ax is)
Thus, the strength of the electric field is 2500 N/C and its direction is horizontal.
Assess: Because the charges are located symmetrically on either side of the y-axis and
are of equal value, the
y-components of their fields will cancel when added.
P20.32. Prepare: The high part of the bottom conductor is closest to the upper conductor, so the positive
charges like to congregate there.
Solve:
(c) We see from the figures that the field is strongest at the high point of the bottom conductor (the earth); this is
where the air is mostly likely to break down and become conducting and allow the opposite charges to rush
toward each other (we call this lightning).
Assess: This is why we are told to stay away from high and tall things in a lightning
storm. Stay away from tall things and crouch down low and hug your knees if you must
remain outside. Inside metal cars is usually a safe place to be.
P20.49. Prepare: The charges are point charges. Please refer to Figure P20.49. Placing the 1 nC charge at
the origin and calling it q1, the q2 charge is in the first quadrant, the q3 charge is in the fourth quadrant, the q4
charge is in the third quadrant, and the q5 charge is in the second quadrant. The electric force on q1 is the vector
sum of the forces F2 on 1, F3 on 1, F4 on 1, and F5 on 1.
Solve:
So,
The magnitude of the four forces is the same because all four charges are equal and equidistant from q1.
9
F2 on 1 = F3 on 1 = F4 on 1 = F5 on 1 =
2
−9
−9
(9.0 × 10 N m2/C )(2 × 10 C)(1 × 10 C)
−4
= 3.6 × 10 N
−2
2
−2
2
(0.5 × 10 m) + (0.5 × 10 m)
Thus, Fon 1 = (3.6 × 10−4 N, toward q2) + (3.6 × 10−4 N, toward q3) + (3.6 × 10−4 N, toward q4) + (3.6 × 10−4 N,
toward q5). It is now easy to see that the net force on q1 is zero.
Assess: Look at the symmetry of the charges. It is
q1 is zero.
no surprise that the net force on charge
P20.51. Prepare: The charges are point charges. Please refer to Figure P20.51. Placing the 1 nC charge at
the origin and calling it q1, the −6 nC is q3, the q2 charge is in the first quadrant, and the q4 charge is in the second
quadrant. The net electric force on q1 is the vector sum of the electric forces from the other three charges q2, q3, and
q4.
Solve: We have
F2 on 1 =
K | q1|| q2|
, away fro m q2
r2
9
=
(9.0 × 10 N m2/C 2)(1 ×10 −9 C)(2 ×10−9 C)
, away from q2
(5.0 × 10−2 m)2
= (0 .72 × 10 −5 N, away from q2 )
r
K | q1|| q3|
F3 on 1 =
, toward q3
r2
=
(9.0 × 109 N m2/C 2)(1 ×10 −9 C)(6 × 10−9 C)
, toward q3
(5.0 × 10−2 m)2
= (2 .16 ×10 −5 N, away from q3 )
r
K | q1|| q4|
F4 on 1 =
, away from q4 = (0 .72 × 10 −5 N, away from q4 )
r2
r
( F2 o n 1 )x = − (0 .72 × 10 −5 N)( cos4 5°) = −( 0.50 9 × 10 −5 N)
r
( F2 o n 1 )y = − (0.72 × 10 −5 N)(sin 45°) = − (0.509 × 10 −5 N)
r
( F3 o n 1 ) x = 0 N
r
( F3 o n 1 ) y = (2.16 × 10 −5 N)
r
−5
−5
( F4 o n 1 )x = (0.72 × 10 N)(cos 45°) = ( 0.509 × 10 N)
r
−5
−5
( F4 o n 1 )y = − (0.72 × 10 N)( sin 45°) = − ( 0.509 × 10 N)
r
r
r
r
( Fo n 1 ) x = ( F2 o n 1 )x + (F3 o n 1 ) x + ( F4 on 1 )x = 0 N
r
r
r
r
( Fo n 1 ) y = ( F2 o n 1 )y + (F3 o n 1 ) y + (F4 on 1 ) y = 1.14 × 10 −5 N
So the force on the 1 nC charge is 1.14 × 10–5 N directed vertically up.
P20.57. Prepare: The electron and the proton are point charges. The electric Coulomb force between the
electron and the proton provides the centripetal acceleration for the electron’s circular motion. ω = 2 f.
Solve:
K (e)(e) mv2
2
=
= mrω 2 = mr(2 π f )
r2
r
f =
1
2π
K e2
mr
3
=
(9.0 × 10 9 N m 2 /C 2 )(1.60 × 10 −1 9 C) 2
(9.11 × 10
−3 1
k g)(5.3 × 10
−1 1 3
)
=
1
( 4.12 × 10 16 rev/s) = 6.6 × 101 5 rev/s
2π
Assess: This is an extremely fast motion, nearly 7 × 1015 rpm!
P20.62. Prepare: The electric field in a region of space between the plates of a parallel-plate capacitor is
uniform.
Solve: The electric field inside a capacitor is E = Q/ε0 A. Thus, the charge needed to produce a field of strength
E is
Q = ε 0 AE = (8.85 × 10
−12
C /N m )[π (0.02 m) ](3 × 10 N/C) = 33.4 nC
2
2
2
6
The number of electrons transferred from one plate to the other is
33.4 × 10 −9 C
11
= 2.1 × 10
1.60 × 10−19 C
P20.65. Prepare: The charged ball attached to the string is a point charge. The ball is in static equilibrium in
the external electric field when the string makes an angle θ = 20° with the vertical. The three forces acting on the
charged ball are the electric force due to the field, the weight of the ball, and the tension force.
Solve:
r
r
r
r
In static equilibrium, Newton’s second law for the ball is Fnet = T + w + Fe = 0. In component form,
(Fnet )x = Tx + 0 N + qE = 0 N
(Fnet )y = Ty − mg + 0 N = 0 N
The above two equations simplify to
T sin θ = qE
T cosθ = mg
Dividing both equations, we get
tan θ =
qE
mg
q=
mg tan θ (5.0 × 10−3 kg)(9.8 N/kg) tan2 0°
−7
=
= 1.78 × 10 C = 180 nC
E
100,000 N/C
to two significant figures.
Q21.17. Reason: The electric potential energy of a capacitor is given by Equation 21.22:
1
2
UC = C(∆VC)
2
We apply this equation to each case:
2
1
C ∆VC = (U C )1
2
1 1
2
(U C ) 2 =
C 2 ∆V C = 2(U C )1
2 2
(U C )1 =
(
)
(
)
2
(U C )3 =
1
1
1
(2C ) ∆VC = (U C )1
2
2
2
(U C )4 =
1
1
(4C ) ∆VC
2
3
2
=
4
(U )
9 C 1
Therefore
(UC )2 > (UC )1 > (UC )3 > (UC) 4
Assess: Again we see the squared term dominates. The potential difference across a
capacitor strongly influences the potential energy stored in the capacitor. Real capacitors
are rated according to the voltage they are designed to handle, and putting more voltage
across a capacitor than that for which it is rated can ruin it.
P21.8. Prepare: Energy is conserved. The potential energy is determined by the electric potential. The
figure shows a before-and-after pictorial representation of a proton moving through a potential difference.
Solve: (a) Because the proton is a positive charge and it slows down as it travels, it must be moving from a region
of lower potential to a region of higher potential.
(b) Using the conservation of energy equation,
Kf + U f = K i + U i
Kf + qVf = K i + qV i
Vf − V i =
1
1 1 2
(K i − K f ) =
mv i − 0 J
q
( e) 2
2
∆V =
mv i
(1.6 7 × 10 −2 7 kg)(8.0 × 10 5 m/s)2
=
= 334 0 V = 3000 V
−1 9
2e
2(1.60 × 10 C)
(c)
Ki =
1 2
mv i = q∆V = (e)(3340 V) = 3000 eV
2
Assess: A positive ∆V confirms that the proton moves into a higher potential region.
P21.27. Prepare: We are thinking of a two-plate parallel capacitor. We will use Equation 21.19 with A = L ×
L.
Solve:
From Equation 21.19, the capacitance is
C=
ε0 A
d
=
ε0 L2
d
Cd
L=
=
ε0
(100 × 10−12 F)(0.20 ×10 −3 m)
= 0.0475 m = 4.8 cm
8.85 × 10−12 C2/N m2
Assess: The capacitance depends only on the geometry of the capacitor.
P21.38. Prepare: The capacitance of a capacitor is a measure of the amount of charge that can be stored on
the plates of the capacitor per unit of electric potential difference across the plates. (C = Q/V). The energy stored
2
by a capacitor of capacitance, C with a potential difference V across the plates is U = CV / 2. Note the issue of
the pair of capacitors has never been resolved—this solution disregards how they are connected and treats them
separately.
Solve: (a) The charge stored on each plate of the capacitor is
−5
−2
3
Q = CV = (10 F)(1.7 ×10 V) = 1.7 ×1 0 C
(b) The energy stored in each capacitor is
−5
2
3
2
U = CV /2 = (10 F)(1.7 ×10 V) /2 = 1.4 J
Assess: These values are reasonable.
P21.46. Prepare: Let the unknown charges be Q1 and Q2, then Q1 + Q2 = 30 × 10−9 C. We will use Equation
21.8 for the electric potential energy and solve for the two charges.
Solve:
U=
1 Q1Q2
1
Q1Q2
−6
=
= −180 × 10 J
4πε 0 r12
4πε0 2.0 ×10−2 m
Q1Q2 =
−(180 × 10−6 J)(2.0 × 10−2 m)
−16 2
= −4 .0 × 10 C
9.0 ×109 N m2/C 2
Solving the first equation for Q2 and substituting into the second equation,
−9
Q1 (30 × 10 C − Q 1 ) = − 4.0 × 10
−1 6
C
2
2
−9
Q 1 − (30 × 10 C)Q1 − (4.0 × 10
−9
Q1 =
−9
2
−16
( 30 × 10 C) ± (−30 × 10 C) + 4(4.0 × 10
That is, the two charges are −10 nC and 40 nC.
2
−1 6
2
C )= 0
2
C )
Q1 = + 40 nC and − 10 nC
Assess: As they must, the two charges when added yield a total charge of 30 nC, and
when substituted into the potential energy equation yield U = −1 80 × 10−6 J.
P21.48. Prepare: The net potential is the sum of the scalar potentials due to each charge. Let the point on
the x-axis where the electric potential is zero be at a distance x from the origin. At this point, V1 + V2 = 0 V.
Solve:
Using Equation 21.9:
1
3 .0 × 10 −9 C
4 πε 0
x
+
−1.0 × 10 −9 C
x − 4.0 cm
=0V
− x + 3 x − 4.0 cm = 0 cm
Either −x + 3(x − 4.0 cm) = 0 cm, or −x + 3(4.0 cm − x) = 0 cm. In the first case, x = 6.0 cm and, in the second
case, x = 3 cm. That is, we have two points on the x-axis where the potential is zero.
Assess: Both values seem reasonable.
P21.54. Prepare: Please refer to Figure P21.54. The net potential at the dot is the sum of the potentials due
to each charge. The dot is equidistant from the three charges. We will denote the 2 nC charge as 1, left –1 nC
charge as 2, and the right –1 nC charge as 3. From the geometry in the figure,
1.5 cm 1.5 cm 1.5 cm
=
=
= cos 30°
r2
r3
r1
Solve:
V=
r1 = r2 = r3 =
1.5 cm
= 1.732 cm
cos 30°
The potential at the dot is
1 q1
1 q2
1 q3
2.0 ×10 −9 C 1.0 ×10 −9 C 1.0 ×10 −9 C
9
2
2
+
+
= (9.0 ×10 N m /C )
−
−
= 0V
4πε0 r1 4πε 0 r2 4πε 0 r3
0.017 32 m 0.01732 m 0.01732 m
Assess: Potential is a scalar quantity, so we found the net potential by adding three
scalar quantities.
P21.56. Prepare: Please refer to Figure P21.56. The proton at point A is at a potential of 30 V and its speed
is 50,000 m/s. At point B, the proton is at a potential of −10 V and we are asked to find its speed. Clearly, the
proton moves into a lower potential region, so its speed will increase. Energy is conserved. Also note that the
potential energy is determined by the electric potential.
Solve: The conservation of energy equation Kf + Uf = Ki + Ui is
1 2
1
mvf + (+e)(−1 0 V) = m(50 ,000 m/s)2 + (+e)(30 V)
2
2
2(1.60 × 10−19 C)(40 V)
2
5
v f = (50,000 m/s) +
= 1.0 ×10 m/s
1.67 ×10 −27 kg
Assess: The speed of the proton is higher, as expected.
P21.62. Prepare: The mechanical energy of the electron is conserved. A parallel plate capacitor has a
uniform electric field. The figure shows the before-and-after visual overview. The electron has an initial speed
vi = 0 m/s and a final speed vf after traveling a distance d = 1.0 mm. The electron loses potential energy and gains
kinetic energy as it moves toward the positive plate.
Solve:
The change in the potential energy of the electron is
∆Ue = Uf – Ui = q∆V = (–e)Ed
The change in the kinetic energy of the electron is
1 2 1 2 1 2
∆K = K f − Ki = mv f − mvi = mv f
2
2
2
Now, the law of conservation of mechanical energy gives ∆K + ∆U = 0 J. This means
1
2
2
mvf − e Ed = 0 J
vf =
2eEd
m
=
2(1.60 × 10 −1 9 C )( 20,000 V/m )(1.0 × 10 −3 m )
9.11 × 10 −3 1 kg
6
= 2.7 × 10 m/s
Assess: Note that ∆V = +Ed because the final potential is higher than the initial
potential.
P21.73. Prepare: When the equipotential lines are widely spaced, as at A, it means the potential isn’t
changing as quickly. From Equation 21.16, E = (∆V)/d , this means the field is weaker there.
Solve: (a) Even though they are on the same equipotential line, the electric field strength at A is smaller than
the electric field strength at B because the equipotential lines are farther apart at A.
(b) The same argument applies here. Since the equipotential lines are farther apart at C than at D, then the
electric field strength at point C is smaller than the field strength at D.
(c) We can tackle the direction first. We know that the electric field lines are perpendicular to the equipotential
lines and that they point in the direction of decreasing potential. So the field line at D points in the positive
x-direction.
In order to get a feel for the spacing of the equipotential lines, imagine one more at 3 V going perpendicularly
down through the origin. Along the x-axis the equipotential lines are quite evenly spaced. Then we can calculate
the potential difference (by counting equipotential lines or by subtracting the last from the first) from the point
(0 cm, 0 cm) to the point (1 cm, 0 cm). It looks like about a 5.5 V potential difference between those two points,
so E = 5.5 V/cm, in the positive x-direction.
Assess: In part (b) it does not matter that point C is not on a drawn equipotential line. You could estimate its
potential as −1.5 V, halfway between the adjacent equipotential lines, but even that is not needed to answer the
question.
r
In part (c) a positive little test charge at point D would feel a force equal to F = qE, so it
would feel a force to the right, and that would be rolling down the hill, as we expect.