a free ball of mass 10 gram carries a charge of 5 into 10 to the power minus 8 coulomb what must be the nature and magnitude of charge that should be given to another ball which is at a distance 7 cm below the former ball so that the two balls remain at rest
Answers
Compare the electrostatic force between an electron and proton separated by 0.530 × 10−10 m with the gravitational force between them. This distance is their average separation in a hydrogen atom.
Strategy
To compare the two forces, we first compute the electrostatic force using Coulomb’s law,
F
=
k
∣
q
1
q
2
∣
r
2
F=k∣q1q2∣r2. We then calculate the gravitational force using Newton’s universal law of gravitation. Finally, we take a ratio to see how the forces compare in magnitude.
Solution
Entering the given and known information about the charges and separation of the electron and proton into the expression of Coulomb’s law yields
F
=
k
∣
q
1
q
2
∣
r
2
=
8.99
×
10
9
N
⋅
m
2
C
2
×
(
1.60
×
10
−
19
C
)
(
1.60
×
10
−
19
C
)
(
0.530
×
10
−
10
m
)
2
F=k∣q1q2∣r2 =8.99×109N⋅m2C2×(1.60×10−19 C)(1.60×10−19 C)(0.530×10−10 m)2
Thus the Coulomb force is F = 8.19 × 10−8 N.
The charges are opposite in sign, so this is an attractive force. This is a very large force for an electron—it would cause an acceleration of 8.99 × 1022 m/s2 (verification is left as an end-of-section problem).The gravitational force is given by Newton’s law of gravitation as:
F
G
=
G
m
M
r
2
FG=GmMr2,
where G = 6.67 × 10−11 N · m2/kg2. Here m and M represent the electron and proton masses, which can be found in the appendices. Entering values for the knowns yields
F
G
=
(
6.67
×
10
−
11
N
⋅
m
2
/kg
2
)
×
(
9.11
×
10
−
31
kg
)
(
1.67
×
10
−
27
kg
)
(
0.530
×
10
−
10
m
)
2
=
3.61
×
10
−
47
N
FG=(6.67×10−11N⋅m2/kg2)×(9.11×10−31 kg)(1.67×10−27 kg)(0.530×10−10 m)2=3.61×10−47 N
This is also an attractive force, although it is traditionally shown as positive since gravitational force is always attractive. The ratio of the magnitude of the electrostatic force to gravitational force in this case is, thus,
F
F
G
=
2.27
×
10
39
FFG=2.27×1039.
Discussion
This is a remarkably large ratio! Note that this will be the ratio of electrostatic force to gravitational force for an electron and a proton at any distance (taking the ratio before entering numerical values shows that the distance cancels). This ratio gives some indication of just how much larger the Coulomb force is than the gravitational force between two of the most common particles in nature.