# Giancoli Physics (5th ed) Chapter 16

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Main Page > Giancoli Physics (5th ed) Solutions > Giancoli Physics (5th ed) Chapter 16

## Problems

### 1.How many electrons make up a charge of -30.0 $\mu$C?

Since electrons have charges of –1.602 x 10-19 C all we need to do is divide the given charge by the electron charge to give us the number of electrons. Also, be careful with the whole micro-coulombs thing.
-30.0 $\mu$C = -30.0 x 10-6 C
(-30.0 x 10-6 C) /(-1.602 x 10-19 C) = 1.88 E14 electrons

(2.7 N)

### 3. Two charged balls are 20.0 cm apart. They are moved, and the force on each of them is found to have tripled. How far apart are they now?

Solving this problem is nice, because in this case we can use the formula $F = \frac{kq_1q_2}{r^2}$ We can set two equations equal to one another. $\frac{3kq_1q_2}{r_1^2} = \frac{kq_1q_2}{r_2^2}$
Where r1 is 0.20 m, and we are trying to solve for r2
If we divide both sides by kq1q2 and plug in for r1 we get: $\frac{3}{0.20^2} = \frac{1}{r_2^2}$
Finally,
r2 = .012 m or 11.5 cm

(0.500 N)

### 5.What is the magnitude of the electric force of attraction between an iron nucleus (q=+26e) and its innermost electron if the distance between them is 1.5 x 10-12 m?

This is a straight up F = kq1q2/r2 problem from the hood. It’s not too bad, the only trick here is the nucleus charge and to solve that problem we simply multiply the first charge by +26 e. Plug in the values you must, young Jedi.

F = kq1q2/r2

F = (8.99E9) * (26 * 1.602E-19 * 1.602E-19) / (1.5E-12)2

F = .0027 N or 2.7E-3 N

(9.2 N)

### 7.What is the magnitude of the force a +15 - mC charge exerts on a +3.0-mili coulombs charge 40 cm away? (1 $\mu$C=10-6 C, 1 mC=10-3C.)?

Questions like these are ones you cross your fingers for on the test. Watch out for the whole mili- and mico- coulomb thing and remember that two positive charges repel. Other than that you can pound this question out with speed.

F = kq1q2/r2

F = (8.99E9) * (15E-6) * (3E-3) / (.4)2

F = 2.5E3 N

### 8.

(3.8x1014 electrons, 3.4x10-16 kg)

### 9. Imagine space invaders could deposit extra electrons in equal amounts on the earth and now your car, which has mass of 1050 kg. note that the rubber tires would provide some insulation. How much charge Q would need to be placed on your car (same amount on earth) in order to levitate it (overcome gravity) (hint : assume that earth charge is spread uniformly so it acts as if it were located at the earth's center, and then the separation distance is the radius of the earth.)?

Ok, seriously man. The author had a little too much time on his hands to write this problem. Buts it’s assigned so lets get to it.
Alright, so the first step is to figure out exactly how much force is required to move your car. Using gravity as the acceleration to counteract, remember these guys are levitating your car, we can use F = ma.

F = (1050 kg) (9.8 ms-2) = 10,290 N

Now, your car is on the surface of the earth, so the center to center distance is the radius of the earth essentially, or 6.38 x 106 m.; Setting the coulombic repulsion equal to the force of gravity on the car we have:

10,290 N = kq1q2/r2

Now, since the two charges must be the same (read the problem), this reduces to:

10,290 N = kq2/r2

10,290 N = (8.99 x 109 Nm2/C2)q2/(6.38x106m)2

q = 6825.7 C = 6.8e3 C

(-5.4x107 C)

### 11. Particles of charge +70, +48, and +80 mC are placed in a line (fig. 16-37) the center one is 0.35m from each of the others. Calculate the net force on each charge due to the other two.

First of all, you should have already sketched a diagram of

the figure because it makes things much, much easier. This problem really
isn’t that difficult, because we don’t have to deal with angles and instead
it is
just plain tedious. Solving this requires you to calculate both forces exerted
on each particle, determining whether that force is either attractive (like
me) or repulsive (like my stupid jokes). I will show how to use all the
necessary calculations for the +70 charge and after that, only the calculated
values.

(a) The force on the; +70E-6:;

Using F =
KQ1Q2/r2

+70 and +48:

F = kq1q2/r2
=
(8.99E9)(70E-6)(48E-6) / (.35)2 = 246.6 N (left - the 48 repels the 70)

+70 and –80:

F = kq1q2/r2
=
(8.99E9)(70E-6)(-80E-6) / (.7)2 = 102.7 N (right - attracted)

Then, just do a little net force action:

102.7 (right) - 246.6 (left) = -144 N
(left)

(b) The force on the middle one (The 48E-6 C)

Using F =
KQ1Q2/r2

+48 and +70 :

F = kq1q2/r2
= (8.99E9)(48E-6)(70E-6) / (.35)2 = 246.6 N (right - the 70 repels the 48)

+48 and –80:

F = kq1q2/r2
= (8.99E9)(48E-6)(80E-6) / (.35)2 = 281.8 N (right - attracted)

Then, just do a little net force action:

246.6 N (right) + 281.8 N (right) = 528 N (right)

(c) The force on the right one (The -80E-6 C)

Using F =
KQ1Q2/r2

-80 and +48 :

F = kq1q2/r2
= (8.99E9)(80E-6)(48E-6) / (.35)2 = 281.8 N (left - attracted

-80 and +70:

F = kq1q2/r2
= (8.99E9)(80E-6)(70E-6) / (.70)2 = 102.7 N (left - attracted)

Then, just do a little net force action:

281.8 N (left) + 102.7 N (left) = 385 N (left)

### 12.Three positive particles of charge 11.0 $\mu$C are located at the corners of an equilateral triangle of side 15.0 cm (Fig. 16-38).; Calculate the magnitude and direction of the net force on each particle.

This problem seems hard because it seems like we’re going to

end up finding three different answers and plus, they’re not in a straight
line. However, all we need to do is calculate the magnitude and net force on
one particle because we are dealing with an equilateral triangle. To solve the
problem, we need to find the force one particle exerts on the
other.

F = kq1q2/r2 =

(8.99E9)(11E-6)2 / (.15)2 = 48.35 N

Now, because this is an equilateral triangle, we can assume

the angle that each particle is exerting on the other is 60o. Since the x-axis
displacement of each particle is the opposite of the other, they cancel one
another out and all we need to know is the y-axis displacement. To find it we
use good old Mr. Sine.

(48.35)(sin 60) = 41.87 * 2 (remember there are two different

particles acting on each) =

83.7 N (away from the center)

### 13. A charge of 6.00 mC is placed at each corner of a square 1.00 m on a side.; Determine the magnitude and direction of the force on each charge.

Make sure you make a diagram and draw arrows as to where the force is going. Use F = kq1q2/r2 and calculate the magnitude and impact each (meaning all 3) charge has on the other.
First find the force of the two particles that are not on a diagonal. Since each particle produces and equal force, the charge will be propelled diagonally, at an angle of 45 degrees.;
F = kq1q2/r2 = (8.99E9)(6E-3)2; / (1)2 = 323640 N
3236402 +3236402; = c2 ;

c = 457696 N

Now, we solve for the third charge on the diagonal.

F = kq1q2/r2 = (8.99E9)(6E-3)2; / 2

F = 161820 N
Now just add the two and the answer is yours.
F = 161820 + 457696 = 619516 N = 6.2 E5 N

### 14.

(2.96x105 N Toward the center of the square)

### 15.Compare the electric force holding the electron in orbit around the proton nucleus of the hydrogen atom, with the gravitational force between the same electron and proton.; What is the ratio of the these two forces?

Solving this require us to use F = kq1q2/r2 to calculate the electric force holding the electron in addition to the gravitational force between the two.
After calculating each, to find the ratio we simply divide the electric force by the gravitational. Use the proton and electron’s charge for q1 and q2, the radius is diameter of a hydrogen atom divided by two.

Electric Force: F = kq1q2/r2 = (8.99E9)(-1.602E-19)(1.602E-19) / (.53E-10)2 = 8.2E-8 N

Gravitational Force: F = Gm1m2/r2 = (6.67e-11)(1.673e-27)(9.11e-31) / (.53e-10)2 = 3.6E-47 N

Divide and Conquer: 8.2E-8 / 3.6E-47 = 2.3E39 N ratio Electric/Gravitational

(5.71 x 1013 C)

### 17.

(Q1 = 1/2QT and Q1, Q2 = 0)

(2.1x10-10 m)

### 19. A + 5.7 $\mu$C and a - 3.5 $\mu$C charge are placed 25 cm apart.; Where can a third charge be placed so that it experiences no net force?

<P>So we're trying to find r. We want the Forces to be equal, so:</P> <P>kq1q2/r2 = kq1q2/r2 </P> <P>Q1 is the "third charge," so we can just cancel it out here, as well as the K. So then we're left with:</P> <P>(3.5 mC)/r2 = (5.7 mC)/(r+.25)2</P> <P>Square root top and bottom on both sides...</P> <P>(3.5 mC)^(1/2)/r = (5.7 mC)^(1/2)/(r+.25)</P> <P>((3.5 mC)^(1/2))r + ((3.5 mC)^(1/2))(.25) = ((5.7 mC)^(1/2))r</P> <P>Solving for r, we get:</P> <P>r = .91 m beyond the negative charge</P>

### 20.

(50.0x10-6 C, 30.0x10-6 C or -15.7x10-6 C, 95.7x10-6 C)

### 21.

The acceleration is produced by the force from the electric field:

F = qE= ma;

(1.60x10-19C)(600N/C) = (9.11x10-31kg)a, which gives a= 1.05x1014 m/s2

Because the charge on the electron is negative, the direction of force, and thus the acceleration, is opposite to the direction of the electric field

The direction of the acceleration is independent of the velocity

### 22.

If we take the positive direction to the east, we have

F = qE = (-1.60x10-19C)(+3500N/C) = -5.6x10-16N, or -5.6x10-16N (west)

### 23.

If we take the positive direction to the south, we have

F = qE;

3.2x10-14N = (+1.60x10-19C)E, which gives E =

+2.0x105N/C (south)

### 24.

If we take the positive direction up, we have

F = qE;

+8.4N = (-8.8x10-6C)E, which gives E =

+9.5x10 5N/C (up)

### 25.

The electric field above a positive charge will be away from the charge, or up.

We find the magnitude from

E = kQ/r2

=(9.0x109N*m2/C2)(33.0x10-6C)/(0.020m)2

=3.30x106N/C (up)

### 26.

The directions of the fields are determined from the signs of the charges are are indicated on the diagram. The net electric field will be to the left. We find its magnitude frm

E = kQ1/L2 = k(Q1+Q2)/l2

= (9.0x109N*m2mC2)(8.0x10-6C+6.0x10-6)/(0.020m)2

=3.2x108N/C

Thus the electric field is 3.2X108N/C toward the negative charge

### 27.

The acceleration is produced by the force from the electric field:

F = qE = ma;

(-1.60x10-19C)E = (9.11x10-31kg)(125 m/s2), which gives E = -7.12x10-10N/C

Because the charge on the electron is negative, the direction of force, and thus the acceleration, is opposite to the direction of the electric field, so the electric field is -7.12x10-10N/C south

### 28.

The direction of the fields are determined from the signs of the charges and are in the same direction, as indicated on the diagram. The net electric field will be to the left. We find its magnitude from

E = kQ1/L2+kQ2/L2 = k(Q=Q)/L2 = 2kQ/l2

1750 N/C = 2(9.0x109N*m2/C2)Q/(0.020m)2, which gives

Q = 6.2x10-10C

### 29.

(2kQa/(a2+x2)3/2 parallel to the line of the changes)

### 32.

(4.82x104 N/C away from the opposite corner)

### 33.

a. (1.73kQ/L2 60° below the - x-axis)

b. (kQ/L2 30° below the = x-axis)

(0.10 N/C)

### 36.

(3.42x105 km from the center of the Earth)

(Q2 = 4Q1)

(8.83x106 m/s)

### 39.

(a. in the direction of the velocity, to the right)

(b. 5.1x102 N/C)

(a. 3x10-10 N)

(b. 7x10-10 N)

(c. 10-4 N)

(5.08 m)

(3.2x10-10)

### 43.

(1.02x10-7 N/C (up))

(6.8x105 C)

(Negative)

### 45.

(1.0x107 electrons)

### 46.

(this is on the wrong block of the syllabus - I didn't show you how to do it yet)
(8.1x105 N/C up)

### 47.

(2.4x10-10 N attraction)

### 48.

(A charge of 0.402 Qo, 0.366 L from Qo)

### 49.

(positive, 2.2x10-7 C)

### 50.

(3.6 m from the positive charge, and 1.6 m form the negative charge)

(a. 0.83 mm)

(b. 2.2 ns)

### 52.

(1.08x107 N/C / (3 - cos((12.5s-1)t))2 up)

### 53.

(5.4 $\mu$C)

(5x10-9 C)

### 55.

(+0.19 N (right))

### 56.

(-1.03x107 N/C (down))

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