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MST 589D [LMGL [IVI XLI LEVHIWX X[S ;SYPH PMOI XS
HW#3 Coulomb’s Law Worksheet WII ]SY HS XLIQ
=SY RIIH ZIGXSVW
Worksheet 1 1. Find the force between charges of +10.0 µC and -50.0 µC located 20.0
cm apart.
Given:
𝑄! = 10 𝑥 10!!
𝑄! = 50 𝑥 10!!
𝑑 = 20 𝑐𝑚 = 0.2𝑚
𝐹 = 9𝑥 10!
𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛:
𝑞! 𝑞!
𝐹=𝑘 !
𝑑
(10 𝑥 10!! )(−50 𝑥 10!! )
𝐹 = 9𝑥 10
(0.2)!
9 10 50 𝑥 (10! )(10!! )((10!! )
𝐹 = 0.04
!
𝐹 = −112.5 𝑁
𝐹 = 112.5 𝑁
2. Two spheres have identical charges and are 75 cm apart. The force
between them is +0.30 N. What is the magnitude of the charge on
each sphere? (Let x = charge)
Given:
𝐹 = 0.30𝑁
𝑑 = 75 𝑐𝑚 = 0.75 𝑚
x = charge
Solution:
0.30(0.75)!
= 𝑥!
9𝑥 10!
0.16875
= 𝑥!
9𝑥 10!
𝐹𝑑!
= 𝑞! 𝑞!
𝐾
1.6875 𝑥 10!!
= 𝑥!
9𝑥 10!
𝑥 ! = 0.1875 𝑥 10!! 𝑥 ! = 1.875 𝑥 10!!! 𝑥 = 1.875 𝑥 10!!! 𝑥 = 4 𝑥 10!! 𝑥 = 4𝜇𝐶 What can you tell about the charge signs on the spheres? It Could be both
negatives or both positives.
3. Consider the electric force between a pair of charged particles a
certain distance apart. By Coulomb's Law:
a. If the charge of one of the particles is doubled, the force is(unchanged) (halved) (doubled) (quadrupled)
b. If, instead, the charge of both particles is doubled, the force
is- (unchanged) (halved) (doubled) (quadrupled)
c. If, instead, the distance between the particles is doubled, the
force becomes- (one fourth) (half) (double) (4 times)
d. If the distance is halved and the charges of both particles are
doubled, the force is 4 times as great.
4. What is the distance between two spheres, each with a charge of 2.5
-6
X 10 C, when the force between them is 0.50 N?
Given:
𝑄! = 𝑄! = 2.5 𝑥 10!!
𝐹 = 0.50 𝑁
d= ?
Solution:
𝑑 = 𝑘𝑞! 𝑞!
𝐹
9𝑥 10! 2.5 𝑥 10!! 2.5 𝑥 10!!
0.50
𝑑 = 𝑑=
9 2.5 2.5 𝑥 (10! )(10!! )(10!! )
0.5
𝑑 = 112.5 𝑥10!!
𝑑 = 0.1125
d= 0.33541 m
d = 33.5 cm
Worksheet 2
1. A negative charge of - 2.0 C and a positive charge of 3.0 C are separated by 80 m.
What is the force between the two charges?
9𝑥 10! (−2)(3)
𝐹 = (80)!
54 𝑥 10!
𝐹 = 6.4 𝑥 10!
𝐹 = 8.4375 𝑥 10! 𝑁
𝐹 = 8.4375 𝑥 10! 𝑁
2. A negative charge of - 0.0005 C exerts an attractive force of 9.0 N on a second
charge that is 10 m away. What is the magnitude of the second charge?
𝐹𝑑 !
𝑞! = 𝑘𝑞!
9(10)!
𝑞! = (9𝑥 10! )(5 𝑥 10!! )
9𝑥10!
𝑞! =
(9𝑥 10! )(5 𝑥 10!! )
1
𝑞! = 𝑥 10!!
5
𝑞! = 0.2 𝑥 10!!
𝑞! = 2 𝑥 10!!
𝑞! = 0.0002 𝐶
3. Two negative charges that are both - 3.0 C push each other apart with a force of
19.2 N. How far apart are the two charges?
𝑑 = 𝑑 = 𝑘𝑞! 𝑞!
𝐹
9𝑥 10! 3 3 19.2
𝑑 = 64951.9 ≈ 64952 𝑚
-5
-5
4. A negative charge of - 4.0 x 10 C and a posivite charge of 7.0 x 10 C are
separated by 0.15 m. What is the force between the two charges?
(4 𝑥 10!! )(7 𝑥 10!! )
!
𝐹 = 9𝑥 10
(0.15)!
!
9 4 7 𝑥10 ( 10!! )( 10!! ) 𝐹 = 2.25 𝑥 10!!
𝐹 = 112 𝑥 10!
𝐹 = 1120 𝑁
-6
5. A negative charge of - 8.0 x 10 C exerts an attractive force of 12 N on a second
charge that is 0.050 m away. What is the magnitude of the second charge?
𝐹𝑑 !
𝑞! = 𝑘𝑞!
12(0.050)!
𝑞! = (9𝑥 10! )(8 𝑥 10!! )
𝑞! =
0.03
72 𝑥10!
𝑞! = 4 𝑥10!! 𝐶
-5
6. Two negative charges that are both - 5.0 x 10 C push each other apart with a
force of 15 N. How far apart are the two charges?
𝑑 = 𝑘𝑞! 𝑞!
𝐹
𝑑 = 9𝑥 10! 5𝑥 10!! 5𝑥 10!! 15
𝑑 = 9 5 (5)
𝑥 (10! )(10!! )(10!! )
15
𝑑=
15 𝑥 10!!
𝑑 = 1.5
𝑑 = 1.22 𝑚
Unit 1Worksheet 3
q1q2
, where r2
2
k = 9.0 × 109 Nm 2 , describe in words the relationship among electric force, C
charge, and distance. The electric force is proportional to the product of the charges and is inversely proportional to the square of the distance between the charges. 2. By how much does the electric force between a pair of charged bodies diminish
when their separation is doubled? tripled?
1. Given the mathematical representation of Coulomb’s Law, F = k
𝐹! =
𝑘
𝑘
𝑘
1
𝐹! = = ! 𝐹! = 𝐹!
!
!
𝑟
(2𝑟)
4𝑟
4
The force is ¼th as much 𝐹! =
𝑘
𝑘
𝑘
1
𝐹! = = ! 𝐹! = 𝐹!
!
!
𝑟
(3𝑟)
9𝑟
9
The force is 1/9th as much. 3. Two positive charges of 6.0 x 10-6 C are separated by 0.50 m. Draw a force
diagram for each of the charges, considering only electrostatic forces. What is the
magnitude of the force between the charges? Is this force repulsive or attractive?
F +6 F +6 0.5
m 𝑞! 𝑞!
𝑑!
9𝑥 10! (6𝑥 10!! )(6𝑥 10!! )
𝐹 = (0.50)!
𝐹 = 1.3 𝑁
Like charge repels, so the force is repulsive. 𝐹=𝑘
4. A negative charge of 2.0 x 10-4 C and a positive charge of 8.0 x 10-4 C are
separated by 0.30 m. What is the magnitude of the force between the charges? Is
this force repulsive or attractive?
𝑞! 𝑞!
𝑑!
9𝑥 10! (2𝑥 10!! )(8𝑥 10!! )
= (0.30)!
9 2 8 𝑥 10!
=
9 𝑥10!!
= 16 𝑥 10!
= 1.6 𝑥 10!
= 16000 𝑁
𝐹=𝑘
𝐹
𝐹
𝐹
𝐹
𝐹
Opposite charges attract, so the force is attractive. 5. A young man accumulates a charge q1 of +2.0 x 10-5 C while sliding out of the
front seat of a car. His girlfriend, who had been waiting in the wind, has picked up
some extra electrons and now has a charge q2 of -8.0 x 10-5 C. Draw a sketch of
the situation. Estimate the magnitude of the electrical force that each person
exerts on the other when separated by a distance of 6.0 m. Is the force attractive or
repulsive?
+ F F 6m -­‐ 𝑞! 𝑞!
𝑑!
9𝑥 10! (2𝑥 10!! )(8𝑥 10!! )
𝐹 = (6)!
9 2 8 𝑥 10!!
𝐹=
36
𝐹 = 4 𝑥 10!!
𝐹 = 0.4 𝑁
Opposite charges attract, so the force is attractive. 𝐹=𝑘
6. Suppose the two people in the previous problem move toward each other.
Calculate the magnitude of the electrical force of one on the other when their
separation is reduced by a factor of 10.
𝑞! 𝑞!
𝑑!
9𝑥 10! (2𝑥 10!! )(8𝑥 10!! )
𝐹 = (0.6)!
9 2 8 𝑥 10!!
𝐹=
0.36
𝐹 = 400 𝑥 10!!
𝐹 = 40 𝑁
7. The figure below shows three point charges that lie along the x axis. Determine
the magnitude and direction of the net electrostatic force on charge q1.
𝐹=𝑘
! !
𝐹 = 𝑘 !! !! 9𝑥 10! (4𝑥 10!! )(6𝑥 10!! )
𝐹 = (0.20)!
9 4 6 𝑥 10!!
𝐹=
0.04
𝐹 = 5400 𝑥 10!!
𝐹 = 5.4 𝑁 𝑡𝑜 𝑡ℎ𝑒 𝑙𝑒𝑓𝑡
Net Force: (-­‐5.4) + (16.8) = 11.4 to the right. 𝐹=𝑘
𝐹
𝑞! 𝑞!
𝑑!
(9𝑥 10! )(6𝑥 10!! )(7𝑥 10!! )
= (0.15)!
(9)(6)(7)𝑥 10!!
𝐹=
0.0225
𝐹 = 16800 𝑥 10!!
𝐹 = 16.8 𝑁 𝑡𝑜 𝑡ℎ𝑒 𝑟𝑖𝑔ℎ𝑡