PHYS 1112
Chapter 15 Problems

Solutions

1
Since hydrogen has an atomic weight of 1.0 gm per mole and one electron and one proton per atom, then we have one mole of electrons and one mole of protons. The fact that each of these has the same magnitude of charge means that we have the same amount of charge at each pole, i.e.

Q = (1.6 x 10-19 C)(6.02 x 1023) = 9.6 x 104 C.

Since the poles are a distance 1.3 x 107 m apart, this means that the force between the charges is

F = k Q2/r2 = (9.0 x 109 N m2/C2)(9.6 x 104 C)2/(1.3 x 107 m)2 = 4.9 x 105 N.

2
The first thing that we must remember is that the force due to a charge distribution is just the vectoral sum of the forces between the individual charges. This means that we must calculate the force between the 3 nC and 6 nC charges, then the force between the 2 nC and the 6 nC, and add them together. The distance between both sets of charges is

d = ((0.5 m)2 + (0.5 m)2)1/2 = .71 m.

For the first pair, the magnitude of the force is

F1 = (9 x 109)(3 x 10-9)(6 x 10-9)/(.71)2 N = 3.2 x 10-7 N.

The magnitude of the other force is

F2 = (9 x 109)(2 x 10-9)(6 x 10-9)/(.71)2 N = 2.2 x 10-7 N.

Now we need to add these forces VECTORALLY, repeat VECTORALLY (groan, boo, hiss, etc.).

To do this, we must write each of the forces in its vector components. For each pair, we note that the forces make a 45 o angle with respect to the x-axis, with the first one being below the axis and the other one above the axis. Therefore, we can write

F1 = F1(cos(-45o ), sin(-45o)) = 3.2 x 10-7 N(.71, -.71) = (2.3 x 10-7 N, -2.3 x 10-7 N).

For the second pair, we have

F2 = 2.2 x 10-7 N(cos(45 o),sin(45o)) = (1.6 x 10-7, 1.6 x 10 -7).

Thus, the total force is

F = F1 + F2 = (3.9 x 10-7 N, -.7 x 10-7 N).

3
If we neglect the interaction between the electron and the proton, then we simply need to calculate the force on each due to the electric field. We have F = qE = ma . Solving for the acceleration, we get a = (q/m)E . Since the proton and electron are oppositely charged, they will accelerate in opposite directions. The magnitude of the proton's acceleration is

ap = (1.6 x 10-19 C)(370 N/C)/(1.7 x 10-27 kg) = 3.5 x 1010 m/s2

and that of the electron's is

ae = (1.6 x 10-19 C)(370 N/C)/(9.1 x 10-31 kg) = 6.5 x 1013 m/s2.

Since the acceleration is constant for both, we know that after a time t, the displacement is given by x = vot + .5 a t2. Therefore, the proton has gone a distance

xp = (.5)(3.5 x 1010)(1.0 x 10-6 )2 = 1.6 x 10-2 m

while the electron has gone

xe = (.5)(6.5 x 1013)(1.0 x 10-6 )2 = 3.3 x 101 m.

Thus, the proton and electron are a distance .016 m + 33 m = 33 m.

4
If we were to draw a force diagram on the foil, then we would have an arrow pointing down representing gravity and an arrow pointing upward representing the tension in the string. Once we turn on an electric field pointing upward, then we will have an additional upward force representing the electrostatic force. Summing the forces, we would get

SFy = T + Fe - Fg = T + qE - mg.

Since the aluminum is not moving, this sum is equal to zero, which means that

T + qE = mg

If the tension in the string goes to zero, then qE = mg. Using the data in the problem, we calculate the electric field to be

E = mg/q = (.0500 kg)(9.80 m/s2)/(3.00 x 10 -6 C) = 1.63 x 105 N/C

5
(a) Using the results of the last problem, we know that the acceleration of the proton is

a = (1.6 x 10-19 C)(640 N/C)/(1.7 x 10-27 kg) = 6.0 x 1010 m/s2.

(b) Since the acceleration is constant, we know that vf = at + vo. Therefore, using vo = 0, we have

t = vf/a = (1.2 x 106 m/s)/(6.0 x 1010 m/s2) = 2.0 x 10-5 s.

(c) The distance that it has gone is given by x = .5 a t2 = (.5)(6.0 x 1010)(2.0 x 10-5 )2 m = 12 m.

(d) The kinetic energy is given by .5 mv2 = (.5)(1.7 x 10-27 kg)(1.2 x 106 m/s) 2 J = 1.2 x 10-15 J.

6
In order to solve this problem, we need to figure out how much force is required to be applied to the ball in order to lift it against gravity the amount stated. To do this, we need to draw a force diagram.

This shows us that we have two equations when the ball is not moving, mg = T cos(15o) and F = qE = T sin(15o). Since mg = (.002 kg)(9.8 m/s2) = .020 N, this means that T = (.020 N)/cos (15o) = .020 N. Therefore,

q = T sin(15o)/E = (.020 N)(.26)/(10 3 N/C) = 5.2 x 10-6 C.

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