Physics 2220 – Module 01 Homework
1.
A glass rod that has been charged to +15.0 nC touches a metal sphere. Afterword, the rod's charge is
+8.00 nC.
(a)
What kind of charged particle was transferred between the rod and the sphere, and in which
direction? That is, did it move from the rod to the sphere or from the sphere to the rod?
The glass rod goes from +15.0 nC to +8.00 nC, thus it lost positive charge. This was done
by gaining electrons. Therefore electrons went from the metal sphere to the glass rod.
(b)
How many charged particles were transferred?
One electron charge is:
e = 1.60 × 10−19 C / electron
The total charge Q is related to the number of charges N (electrons) by:
Q = Ne
Solve for the total number of electrons lost. The total charge difference is 7.00 nC.
N=
2.
Q
7.00 × 10−9 C
=
= 4.38 × 1010 electrons
−19
e
1.60 × 10 C / electron
A plastic balloon that has been rubbed with wool will stick to a wall.
(a)
Can you conclude that the wall is charged? If not, why not? If so, where does the charge come
from?
No, the force of attraction can not only be from oppositely charged particles, but also
from a neutral object that is polarized by a charge.
(b)
3.
Draw a series of charge diagrams showing how the balloon is held to the wall.
Two small plastic spheres each have a mass of 5.0 g and a charge of -55 nC. They are placed 2.0 cm
apart (center to center).
(a)
What is the magnitude of the electric force on each sphere?
Since the plastic spheres are both negatively charged, they will repel each other and the
direction of the forces will be in opposite directions.
The magnitude of force can be found using Coulomb's Law:
∣F 12∣ = ∣F21∣ = F E = k e
∣q 1∣∣q 2∣
d2
−9
C∣∣−55 × 10−9 C∣
9
2
2 ∣−55 × 10
F E = (8.99 × 10 N m /C )
= 6.8 × 10−2 N
2
(0.02 m)
(b)
By what factor is the electric force on a sphere larger than its weight?
FE
F
6.8 × 10−2 N
= E=
= 1.4
F G mg (0.005 kg) (9.80 m/s 2)
4.
What is the net electric force on charge A?
The force on charge A will be the sum of the forces caused by charge B and charge C as per Newton's
Second Law. Opposite charges attract and the same charges repel, so charge B will pull charge A
toward it, and charge C will push charge A away from it.
∑ F⃗A = F⃗BA +
F⃗CA
Since the forces are all along the same direction, the vector notion can be dropped and the net electric
force can easily be determined.
∑ F A = F BA +
F CA
q q
q q
∑ F A = ke B 2 A + ke C 2 A
d BA
d CA
q
q
∑ F A = k e q A d 2B + d 2C
BA
CA
(
∑ F A = ( 8.99 × 10
9
2
2
N m / C ) (1.0 × 10
−9
)
−1.0 × 10−9 C 4.0 × 10−9 C
C)
+
2
2
(0.01 m)
(0.02 m)
(
)
∑ F A = ( 8.99 × 109 N m2 / C2 ) ( 1.0 × 10−9 C) (0 C) = 0 N
5.
What are the strength and direction of the electric field that will balance the weight of a 2.0 g plastic
sphere that has been charged to – 5.0 nC?
Since the sphere is negatively charged, the electric field must go down so FE can be up and balance Fg.
∑ F = FE − FG = 0
F⃗E = ⃗
E q
⃗
F G = m ⃗g
→
→
F E = Eq
F G = mg
Solve the Newton's Second Law equation and find the magnitude of
the electric field.
Eq = mg
2
mg (0.002 kg) (9.80 m/s )
E=
=
= −3.9 × 10 6 N/C
−9
q
−5.0 × 10 C
6.
Objects A and B are both positively charged. Both have a mass of 100 g, but A has twice the charge of
B. When A and B are placed 10 cm apart, B experiences an electric force of 0.45 N.
(a)
What is the charge on A?
1
Q = QB
2 A
The relationship between charge A and B:
Q A = 2Q B
Use Coulomb's Law to solve for the charge on object A:
F = ke
Q A QB
d2
QA
= ke
F = ke
QA =
(b)
√
( 12 Q )
A
d2
Q2A
2d 2
√
2 (0.45 N) ( 0.1 m)2
2Fd 2
=
= 1.0 × 10−6 C
9
2
2
ke
8.99 × 10 N m / C
If the objects are released, what is the initial acceleration of A?
Newton's Third Law says that the force object A will feel is equal and opposite to object B.
F AB = −F BA
F BA = −0.45 N
→
Use Newton's Second Law to find the acceleration.
F AB = maA
aA =
7.
F AB −0.45 N
=
= −4.5 m/s2
m
0.1 kg
What is the force F on the 1.0 nC charge? Give your answer as a magnitude and direction.
Use Newton's Second Law to determine the force on q1. This is in two directions, so x and y
components will need to be used. The direction (positive or negative) will be explicitly placed in the
equations so the absolute values of the charges can be used.
F⃗1 = F⃗21 + F⃗31
In terms of the components:
o
o
F1x = F 21 cos (60 ) + F 31 cos (60 )
o
o
F 1y = F 21 sin (60 ) − F 31 sin (60 )
Now substitute Coulomb's Law into each component and solve for the x-component.
F 1x = k e
∣q2∣∣q1∣
d
2
21
cos (60 o) + k e
F 1x = k e ∣q 1∣ cos (60 o)
(
∣q 3∣∣q 1∣
cos (60o )
2
31
d
∣q2∣
d
F1x = (8.99 × 109 N m2 / C2 ) ∣1.0 × 10−9 C∣ cos (60o )
∣q3∣
+
2
21
(
2
d 31
)
)
∣2.0 × 10−9 C∣ ∣−2.0 × 10−9 C∣
+
2
2
( 0.01 m)
( 0.01 m)
F1x = 1.8 × 10−4 N
Now substitute Coulomb's Law into each component and solve for the y-component.
F 1y = k e
∣q2∣∣q1∣
d
2
21
sin (60 o) − k e
F 1y = k e ∣q 1∣ sin (60 o)
(
∣q3∣∣q1∣
d
∣q2∣
d
−
2
21
F1y = (8.99 × 109 N m2 / C2 ) ∣1.0 × 10−9 C∣ sin (60o )
2
31
(
sin ( 60o)
∣q 3∣
d 231
)
∣2.0 × 10−9 C∣ ∣−2.0 × 10−9 C∣
−
2
2
(0.01 m)
(0.01 m)
)
F 1y = 0 N
Solve for the magnitude and direction:
F1 = √ F 21x + F 21y = √( 1.8 × 10−4 N)2 + (0 N)2 = 1.8 × 10−4 N
θ = tan−1
8.
(
0N
= 0o
−4
1.8 × 10 N
)
Two 10-cm long thin glass rods uniformly charged to +10 nC are placed side by side, 4.0 cm apart.
What are the electric field strengths E1 to E3 at distances 1.0 cm, 2.0 cm and 3.0 cm to the right of the
rod on the left along the line connecting the midpoints of the two rods?
Observations:
• Both rods will contribute to the electric field at each location
• Remember, the charge on both rods is positive and the electric field is a vector:
◦ The part of the electric field contributed by the left rod will point to the right
◦ The part of the electric field contributed by the right rod will point to the left
• The electric field for one rod can mathematically be determined, and then the same reasoning
can be applied to the second rod.
◦ Consider a generic point P from one of the rods:
Assume:
• The rod has a charge of Q
• The rod has a length of ℓ
Reasoning:
• Split the rod into small dq segments, each which contribute a fraction dE to the total electric field
• Each dq segment is related to a length segment dy of the rod by λ, the charge per unit length
dq = λ dy
•
•
Note the symmetry of the locations in question. Any dE contribution in the y-direction from a part
of the rod above the center axis will be canceled out by an dE contribution from the bottom half
of the rod
Therefore, the only net electric field is in the x-direction (Ex)
Start with the equation for the electric field due to a single charge segment:
dE = k e
dq
r2
Only the x-component is needed:
x
x dq
= ke
r
r3
dE x = dE cos ( θ) = dE
Since the rods are along the y-axis, the integration will need to be done over the y-axis. Therefore r and
dq need to be put in terms of y and dy. The relationship between dq and dy has already been shown.
Use Pythagorean Theorem for r and y.
r = √x + y
2
2
Plug this into the electric field equation and integrate:
dE x = k e
x dq
x λ dy
= ke 2
3
r
( x + y 2 )3 / 2
ℓ /2
E x = ke x λ
dy
2 3 /2
−ℓ / 2 (x + y )
∫
2
Solution to the integral can be found in integral tables.
Recall:
λ=
Q
ℓ
Solve for E-field in the x-direction:
[
y
Ex = k e x λ 2 2
x (x + y 2 )1 / 2
Ex = ke
]
ℓ /2
−ℓ / 2
[
Q
ℓ/2
−ℓ / 2
−
2
2 1/ 2
2
ℓ x ( x + (ℓ / 2) )
x (x + (−ℓ / 2)2)1 / 2
ke Q
Ex =
2
x √ x + (ℓ / 2)2
]
This equation can be applied to each position to determine the portion of the magnitude contributed by
each rod to the net electric field at that position. The x is the distance from the left / right rod.
Example, left rod for position 1:
Ex =
(0.01 m) √(0.01 m) + (0.05 m)
2
2
= 1.765 × 105 N/C
Position
Left Rod
Right Rod
Total (left - right)
E1
1.765 × 105 N/C
5.14 × 104 N/C
1.25 × 105 N/C
E2
8.347 × 104 N/C
8.347 × 104 N/C
0 N/C
4
E3
9.
(8.99 × 10 9 N m 2 /C2 ) (10 × 10−9 C)
5.14 × 10 N/C
5
1.765 × 10 N/C
- 1.25 × 105 N/C
Two 10-cm-diameter charged rings face each other, 20 cm apart. Both rings are charged to +20 nC.
What is the electric field strength at (a) the midpoint between the two rings (b) the center of the left ring
In class, the electric field for a ring of radius “a” and distance “x” from the center of the ring was derived:
Ex =
ke x Q
(x 2 + a2 )3 / 2
The charge on both rings is positive, therefore:
• The contribution to the electric field from the left ring will point to the right
• The contribution to the electric field from the right ring will point to the left
•
⃗
E = E⃗left + E⃗right
(a)
the midpoint between the two rings
Contribution of the left ring:
E left =
ke x Q
(8.99 × 10 9 N m 2 /C2 ) (0.10 m) (20 × 10−9 C)
=
(x 2 + a2 )3 / 2
((0.10 m)2 + (0.05 m)2)3 / 2
Eleft = 1.29 × 104 N/C
Contribution of the right ring:
• Because the midpoint is being considered, the magnitude of the contribution will be the
same as the left ring, but it will be in the opposite direction. Therefore:
(b)
the center of the left ring
E midpoint = E left + E rigiht = 0 N/C
Contribution of the left ring:
E left =
ke x Q
(8.99 × 10 9 N m 2 /C2 ) (0 m) (20 × 10−9 C)
=
= 0 N/C
(x 2 + a2 )3 / 2
((0 m)2 + (0.05 m)2)3 / 2
Contribution of the right ring:
ke x Q
(8.99 × 109 N m 2 /C 2) (0.20 m) (20 × 10−9 C)
E right = 2
=
(x + a 2)3 / 2
((0.20 m)2 + (0.05 m)2 )3 / 2
E right = 4.10 × 103 N/C
Total:
10.
Ecenter of left ring = 4.10 × 10 3 N/C
An electron in a uniform electric field increases its speed from 2.0 × 10 7 m/s to 4.0 × 107 m/s over a
distance of 1.2 cm. What is the electric field strength?
The electric field in terms of the force that will be acting on the electron (the test charge) is:
F E = Ee
The force on the electron caused by the electric field is the only force acting on the charge, and will obey
Newton's Second Law.
ma = Ee
ma
E=
e
The mass of an electron (look it up in the book) is 9.11 × 10-31 kg and the charge is given. The
acceleration of the charge needs to be determined. Given the initial and final velocities over a
displacement, the kinematic equations can be used.
2
2
v f = v i + 2a ( Δ x )
2
2
vf − vi
(4.0 × 107 m/s)2 − (2.0 × 107 m/s)2
a=
=
= 5.0 × 1016 m/s2
2 (Δ x )
2 ( 0.012 m)
Now find the electric field.
−31
E=
11.
16
ma (9.11 × 10 kg) (5.0 × 10
=
e
(1.6 × 10−19 C)
2
m/s )
= 2.8 × 105 N/C
The surface charge density on an infinite charged plane is -2.0 × 10 -6 C/m2. A proton is shot straight
away from the plane at 2.0 × 106 m/s. How far does the proton travel before reaching its turning point?
This is the same kind of problem as a projectile mass that is under the influence of gravity, but now it is
positive charge feeling the force of negatively charged surface.
Kinematic equations can be used to determine how far the proton will travel before it reaches the turning
point, where its velocity will be zero.
2
2
v f = v i + 2a (Δ y)
2
−v i
Δ y=
2a
The acceleration is unknown, but can be determined by using what is known about the electric field and
Newton's Second Law. The positive charge (q) is under the influence of the electric field (E) caused by
the surface charge.
F E = ma = Eq
The electric field for an infinite sheet is:
E= σ
2 ϵo
Combine these equations and solve for the acceleration.
ma
E= σ =
2 ϵo
q
qσ
a=
2m ϵo
Now substitute this back into the equation for the distance and solve.
2
Δ y=
12.
2
−vi
−vi m ϵo −(2.0 × 106 )2 (1.67 × 10−27 kg) (8.85 × 10−12 C2 /N m2 )
=
=
−19
−6
2
2a
qσ
(1.60 × 10 C) (−2.0 × 10 C/m )
Δ y = 0.185 m
What are the strength and direction of the electric field at the position indicated by the dot. Give your
answer (a) in component form and (b) as a magnitude and angle measured clock-wise or counter-clockwise (specify which) from the positive x-axis.
Each charge will contribute a part of the net electric field at the point of the dot. An “Electric Field
Diagram” can be drawn similar to a force diagram to show what direction each contribution will point.
Start by solving for the magnitude and direction of each of these contributions.
Charge #1:
∣E⃗1∣ = k e
∣q1∣
r
2
9
2
2
= (8.99 × 10 N m / C )
∣+ 5.0 × 10−9 C∣
2
(0.02 m)
E⃗1 = 112375 N/C ̂i
= 112375 N/C
Charge #2:
∣q2∣
∣−5.0 × 10−9 C∣
9
2
2
⃗
∣E2∣ = k e 2 = (8.99 × 10 N m / C )
2
2 = 22475 N/C
r
(0.02 m ) + (0.04 m )
•
Angle:
θ = tan−1
•
= 63.435
( 42 cm
cm )
o
Components:
∣E 2x∣ = E 2 cos (θ ) = 10051 N/C
∣E 2y∣ = E 2 sin (θ) = 20102 N/C
•
Observe its x-component is negative and y-component is positive
E⃗2 = −E2x ̂i + E 2y ̂j
E⃗2 = (−10051 î + 20102 ̂j) N/C
Charge #3:
−9
∣q3∣
C∣
9
2
2 ∣+ 10.0 × 10
⃗
∣E 3∣ = k e 2 = (8.99 × 10 N m / C )
= 56188 N/C
2
r
(0.04 m )
E⃗3 = −56188 N/C ̂j
(a)
in component form – simply add the components together
E⃗net = E⃗1 + E⃗2 + E⃗3
E⃗net = (112375 ̂i + (−10051 ̂i + 20102 ̂j) + (−56188 ̂j)) N/C
E⃗net = (1.0 × 105 î − 3.6 × 10 4 ̂j) N/C
(b)
as a magnitude and angle measured clock-wise or counter-clock-wise (specify which) from the
positive x-axis
Use Pythagorean Theorem for the magnitude:
∣E⃗net∣ = √(102324)2 + (−36086)2
∣E⃗net∣ = 1.1 × 10 5 N/C
N/C
For the direction, note the x-component is positive the y-component is negative, so the angle will
be below the x-axis, or clockwise from the positive x-axis.
ϕ = tan1
(
36086
= 19.4 o
102324
)