Solutions to Homework Questions 1

Physics 112 Homework 1 (solutions) (2004 Fall)
Chapt15, Problem-7: Suppose that 1.00 g of hydrogen is separated into electrons and protons.
Suppose also that the protons are placed at Earth’s North Pole and the electrons are placed at the South
Pole. What is the resulting compressional force on Earth?
Solution:
1.00 g of hydrogen contains Avogadro’s number of atoms, each containing one proton and one electron. Thus,
each charge has magnitude q = N A e . The distance separating these charges is r = 2 R E , where R E is the
Earth’s radius. Thus applying Coulomb’s Law,
F = k e( N A e ) 2
( 2 R E ) 2
⎛
= 8.99 × 10 9 N ⋅m2 ⎞ ( 6.02 ×10 23
) 1.60 × 10 −19 C
⎜
⎝
C 2 ⎟
⎠ 4 6.38 × 10 6 m
[ ( )] 2
( ) 2 = 5.12 ×105 N
Chapt15, Problem-8: An electron is released a short distance above Earth’s surface. A second
electron directly below it exerts an electrostatic force on the first electron just great enough to cancel the
gravitational force on it. How far below the first electron is the second?
Solution:
The magnitude of the repulsive force between electrons must equal the weight of an electron, Thus by
using Coulomb;’s Law and Newton’s 2 nd Law (applied to gravity), we have
k e e 2 r 2 = m e g
so rearranging this expression, and then making the substitutions, we get
r =
( )( 1.60 ×10 −19 C ) 2
( )( 9.80 m s 2
)
k e e2
m e
g = 8.99 ×10 9 N⋅ m 2 C 2
9.11 ×10 −31 kg
Chapt15, Problem-11: Three charges are arranged
as shown in Figure P15.11. Find the magnitude and direction
of the electrostatic force on the charge at the origin.
Solution:
In the sketch to the right, F R is the resultant of the
forces F 6 and F 3 that are exerted on the charge at
the € origin by the 6.00 € nC and the –3.00 nC charges
respectively. Applying Coulomb’s Law to each, we get
⎛
F 6
= 8.99 ×10 9 N⋅ m 2 ⎞ 6.00 ×10 −9 C
⎜
⎝
C 2 ⎟
⎠ 0.300 m
= 3.00× 10 −6 N
⎛
F 3 = 8.99 ×10 9 N⋅ m 2 ⎞
⎜
⎝
C 2 ⎟
⎠
( )( 5.00 ×10 −9 C )
( ) 2
( 3.00× 10 −9 C) 5.00 × 10 −9 C
( )
0.100 m
From the Superposition Principle, the resultant is
F R = ( F 6 ) 2 ⎛
+ F 3
⎜
⎝
( ) 2 = 1.38 ×10 −5 N at θ = tan −1 F 3
or F R = 1.38 ×10 −5 N at 77.5° below − x axis
= 5.08 m
( ) 2 = 1.35× 10 −5 N
F 6
⎞
⎟ = 77.5° ,
⎠
2