Electric Charges and Coulomb's Law
1. The radius of a hydrogen atom is r = 5.29 × 10
−11
m.
a. What is the ratio of the electric to gravitational force between a proton and an electron in a
hydrogen atom?
A hydrogen atom has one proton in its nucleus and one electron orbiting the nucleus. The
magnitude of the charge of an electron is the same as the magnitude of the charge of a proton
and equals 1.60 × 10−19 C. The magnitude of the electric force is determined by the Coulomb’s
Law:
Fe = k
qe q p
r2
(1.60 × 10 −19 C )(1.60 × 10 −19 C )
= (8.99 × 10 N ⋅ m / C )
= 8.22 × 10 −8 N .
−11
2
(5.29 × 10 m)
9
2
2
Similarly, the magnitude of the gravitational force between the electron and the proton is:
Fg = G
me m p
r2
= (6.67 × 10
−11
(9.11 × 10 −31 kg )(1.673 × 10 −27 kg )
N ⋅ m / kg )
= 3.63 × 10 −47 N
−11
2
(5.29 × 10 m)
2
2
Hence, we obtain the ratio of the two forces:
Fe
8.22 × 10 −8 N
=
= 2.26 × 10 39 = 2,260,000,000,000,000,000,000,000,000,000,000,000,000
− 47
Fg 3.63 × 10 N
b. Find the speed of the orbital motion of an electron.
The electron undergoes a circular motion with a radius r = a 0 =5.29 x 10-11 m. ao is called Bohr
radius. The mass of an electron is me = 9.11 x 10-31 kg
electron
r
Proton
The force acting on the electron:
Fe = k
qe q p
r2
v2
ac is centripetal acceleration. Hence,
By Newton’s 2 Law Fe = mac where ac =
r
nd
k
qe q p
r2
=
mv 2 2 Fr
v =
r
me
v=
Fr
8.22 × 10−8 × 5.29 ×10−11
=
= 2.18 × 106 m / s .
−31
me
9.11× 10
2. Two very small spheres are initially neutral and separated by a distance of 0.50 m. Suppose
that 3.0x1013 electrons are removed from one sphere and placed on the other.
a. What is the magnitude of the electrostatic force that acts on each sphere?
This is an application of Coulomb’s law:
(
(1.602x10
)
C ) = −4.806x10
q1 = +3.0x1013 1.602x10−19 C = +4.806x10 −6 C
q 2 = −3.0x1013
So,
−19
−6
C r=0.5m
2
G
( 4.806x10−6 )
F = k q1rq2 2 = 8.99x109
= 0.831N
2
0.5
( )
By Newton’s law, the force on each sphere is the same.
b. Is the force attractive or repulsive?
The force is attractive because the charges are opposite.
3. The attractive force between two point charges separated by 6 cm is 20 N.
a. What is the electrical force between the charges when they are separated by 12 cm?
By Coulomb’s Law F =
q1q2
r2
The ratio of the forces at the two distances:
q1q2
2
F6
122 144
6
=
= 2 =
=4
6
36
F12 q1q2
122
The force at 6 cm is 4 times as large as at 12 cm. Thus the force at 12 cm is 20N/4 = 5 N
b. If the charges attracting have equal magnitudes, what is the magnitude of each charge?
By Coulomb’s law:
q1q2
q2
k
=
r2
r2
F × r 2 20 × 3.6 × 10−3
q2 =
=
= 8 × 10−12
9
k
9 ×10
F =k
q = 8 × 10−12 = 2.83 × 10−6 C
4. Three positive charges lie on the x-axis; q1 = 25 μC is at the origin, q2 = 10 μC is at x1 = 2 m,
and q3 = 20 μC is a x2 = 3 m. Find the resultant force on q3.
q1
q2
2m
q3
3m
The force on q3 due to q2 is in the positive x direction and has a magnitude of:
F23 = k
q2 q3 (9x10 9 Nm 2 / C 2 )(10x10 -6 C)(20x10 -6 C)
=
= 1.8 N
r23 2
(1 m)2
The force on q3 due to q1, is also in the positive x direction. Its magnitude is:
F13 = k
q1q3 (9x10 9 Nm 2 / C 2 )(25x10 -6 C)(20x10 -6 C)
=
= 0.5 N
r13 2
(3 m) 2
Since both forces are in the positive direction, the resultant is also in that direction and just the
sum of the magnitudes:
F = F13 + F23 = 2.3 N
5. Three charges, each equal to +2.90 μC, are placed at three corners of a square 0.500 m on
each side. Find the magnitude and direction of the net force on charge number 3.
The magnitude of force exerted on charge 3 by charge 1:
y
G
F
G
F31
2
θ
r=0.500 m
G
F32
3
2r
r=0.500 m
x
1
F31 = k
q2
= (8.99 × 10 9 N ⋅ m 2 / C 2 )
( 2r ) 2
( 2.90 × 10 −6 C ) 2
= 0.151 N
2(0.500 m) 2
The magnitude of force exerted on charge 3 by charge 2:
−6
C)2
q2
9
2
2 ( 2.90 × 10
F32 = k 2 = (8.99 × 10 N ⋅ m / C )
= 0.302 N
r
(0.500 m) 2
G
G
The components of F31 and F32 :
F31, x = F31 cos 45.0 o = (0.151 N )(0.707) = 0.107 N
F31, y = F31 sin 45.0 o = (0.151 N )(0.707) = 0.107 N
F = F cos 0 = (0.302 N )(1) = 0.302 N
o
32 , x
32
F32, y = F32 sin 0 o = (0.151 N )(0) = 0 N
The components of the resultant force:
Fx = F31, x + F32, x = 0.107 N + 0.302 N = 0.409 N
Fy = F31, y + F32 , y = 0.107 N + 0 N = 0.107 N
The resultant force acting on charge 3:
F = Fx + Fy = 0.423 N
2
2
The direction of the resultant force on charge 3:
⎛ F3, y ⎞
⎟ = 14.7 o .
⎟
⎝ F3, x ⎠
θ = tan −1 ⎜⎜