Problem Set 16: | 1 | 3 | 5 | 7 | 9 | 11 | 12 | 13 | 15 | 19 | 21 | 23 | 25 | 27 | 30 | 31 | 45 | 51 | Go up
- by D-Fresh Shriver and P-shizzle Pham, Gimps of 2004 - proud to be your copying solution. (Just kidding, that's wrong.)

1. How many electrons make up a charge of -30.0 m

Since electrons have charges of –1.6E-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 mili-coulombs thing.

-30 mC / 1.602E-19 C = 1.88 E14 electrons

3. Two charged balls are 20.0 cm 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 while we use the formula F = kq1q2/r2 we can ignore both charges and the variables F and k. Since the resulting force is tripled, we know that the radius of the new situation will be 1/3 of the original value of r2. To start this problem, we need to set two equations equal to one another. This works because but cutting the original value of r2 by 1/3, the two equations will be equal and we can then solve for r in the new situation.

F = kq1q2/r2

kq1q2/r2 = kq1q2/r2

Remove unnecessary variables and set equations equal by multiplying r2 by 1/3 in equation 1.

1 / ((1/3)* 202) = 1 / r2

(1/3) * (202) = r2

r = .012 m or 11.5 cm

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

7. What is the magnitude of the force a +15 - mC charge exerts on a +3.0-mili coulombs charge 40 cm away? (1 mC=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

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

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 mC 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

F = 161820 + 457696 = 619516 N = 6.2 E5 N

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

19. A + 5.7 mC and a - 3.5 mC charge are placed 25 cm apart.  Where can a third charge be placed so that it experiences no net force? So we're trying to find r. We want the Forces to be equal, so:

kq1q2/r2 = kq1q2/r2

Q1 is the "third charge," so we can just cancel it out here, as well as the K. So then we're left with:

(3.5mC)/r2 = (5.7mC)/(r+.25)2

Square root top and bottom on both sides...

(3.5mC)^(1/2)/r = (5.7mC)^(1/2)/(r+.25)

((3.5mC)^(1/2))r + ((3.5mC)^(1/2))(.25) = ((5.7mC)^(1/2))r

Solving for r, we get:

r = .91 m beyond the negative charge

21. What is the magnitude and direction of the acceleration experienced by an electron in and electric field of 600 N/C?  How does the direction of the acceleration depend on the direction of the field at that point?  How does the direction of the acceleration depend on the electron's velocity at that point?

First, looking at the given information, we know we can find Force using E = F/q. From there, you can easily find the acceleration using F = ma.

Given info:

E = 600 N/C

q = 1.602e-19

E = F/q

600 N/C = F/(1.602e-19)

F = 9.612e-17 N

F = ma

9.612e-17 N = (9.11e-31)a

a = 1.055e14 m/s/s

Because it's an electron, it has a negative charge. This means that the electron should always accelerate in the opposite direction of the E. Field. This is why the direction of acceleration depends on the direction of the E. Field.

23. A proton is released in a uniform electric field, and it experiences an electric force of 3.2 x 10-14 N toward the south.  What are the magnitude and direction of the magnetic field?

Given Info:

F = 3.2e-14

q = 1.602e-19

E = F/q

E = 3.2e-14/1.602e-19

E = 1.997e5 N/C South

Because it is a proton (positive charge), it will always travel with the magnetic field (away from the source is how I think of it), and vice versa. So in this case, the magnetic field is in the same direction as the force on the proton.

25. What is the magnitude and direction of the electric field 30.0 cm directly above a 33.0 x 10-6 C charge?

Now, we're into point charges. In the info given, we have a radius (distance between charge and point charge), as well as 1 Charge (33e-6 C).

E = kq/r2

E = (8.99e9)(33e-6)/(.3)2

E = 3.29e6 Up

Because this charge is positive, the electric field is always away from the charge. If the point is directly above the charge, then the electric field will travel out to the point, and past it... up.

27.  An electron is released from rest in a uniform electric field and accelerates to the north at a rate of 125m/s2. What is the magnitude and direction of the electric field?

Given Info:

a = 125 m/s/s

q = 1.602e-19

In order to solve for E with these givens, we have to use both E = F/q and F = ma. Because you can substitute ma for F in the first equation:

E = F/q = ma/q

E = (9.11e-31)(125)/(1.602e-19)

E = 7.12e-10 south

Electrons always travel against the electric field magnitude. Since the electron is traveling north, the electric field must be south.

30. Use Coulomb's law to determine the magnitude and direction of the electric field at points A and B in Fig. 16-41 due to the two positive charges (Q = 9.0mC) shown. Is your result consistent with Fig. 16-29b? Vector Component Time

We'll solve for A first:

Notice the right triangles that you can create between point A and the Q's.First thing you'll want to do is solve for the hypotenuse on one of the sides. I chose the side on the right (it doesn't matter). A2 + B2 = C2

52 + 102 = C2

C = .0125^(1/2)

The hypotenuse, then, is the distance between A and Q. a.k.a. the radius, r. Since we already know K (a constant, 8.99e9) and Q is given (9mC) we can plug this into our equation for Electric Field:

E = kq/r2

E = (8.99e9)(9mC)/(.0125)

E = 6472800 N/C

Notice that .0125^(1/2) squared is just .0125...

Now that we have the Electric Field, we have to break it down into components. First solve for the angle: Angle is approximately 26.6 degrees. Now use that to find your components.

sin(26.6)*(6472800) = Magnitude Up

cos(26.6)*(6472800) = Magnitude Left

Since the magnitude left will eventually cancel out with the magnitude coming from the Charge on the other side of A, you can ignore that number. And since essentially you'd be doing the same math over again, because its the exact same distance and angle away, the Charge on the left side will yield the same magnitude up. So at this point you can just multiply by two to get the answer:

E = 5.8e8 N/C Up

For B:

I'll be brief with this one. Form your triangles again with B, and this time you'll have to solve for the magnitude from both Charges. Follow the same steps from part A (Hypotenuse, Solve for E, Break down to components, and Add the components). For the right side:

r = .025^(1/2)

E = (8.99e9)(9mC)/(.025)

E = 3236400 N/C

Angle = 18.4 degrees

sin(18.4)*3236400 = 1023439.54 N/C up

cos(18.4)*3235400 = 3070318 N/C left

For the left side

r = .005^(1/2)

E = (8.99e9)(9mC)/(.005)

E = 16182000 N/C

Angle = 45 degrees

sin(45)*16182000 = 11442401 N/C up

sin(45)*16182000 = 11442401 N/C right

1023439 + 11442401 = 12465840 N/C up

11442401 - 3070318 = 8372083 N/C right

(124658402 + 83720832)^(1/2) = 1.5e7 N/C 56 degrees above horiz.

31. Calculate the electric field at the center of a square 60 cm on a side if one corner is occupied by a + 45.0-mC charge and the other three are occupied by - 31.0-mC charges? Since the Charges in the upper right and bottom left actually cancel each other out, we can just focus on the positive charge and the negative charge opposite of it diagonally. Ignore the fact that the negative charges are in fact "Negative", that will only come in handy later when you figure the direction of the magnitude.

r = .18^(1/2)

E = kq/r2

E = (8.99e9)(45mC)/(.18) = 2247500 N/C 45 degrees below horiz

E = (8.99e9)(31mC)/(.18) = 1548278 N/C 45 degrees below horiz

Since the Charge in the bottom left is negative, it is actually pulling instead of pushing that point. Since the push of the positive charge and the pull of the negative charge are in the exact same direction, we just add em:

2247500 + 1548278 = 3.8e6 N/C 45 degrees below horiz.

45. A water droplet of radius 0.018 mm remains stationary in the air.  If the electric field of the Earth is 150 N/C, how many excess electron charges must the water droplet have?

Info Given:

Radius of the droplet: .000018 m

E = 150 N/C

Since there is no equation that can relate these two, we'll have to first use the radius to find mass of the droplet.

m = d(4/3)pr3

d = (density) given = 1000

m = (1000)(4/3)(p)(.000018)3

m = 2.4e-11

Now that we have enough variables, we can use an equation to relate E with m and g. Eq is the electric force, mg is the force of gravity.

Eq = mg

(150)(q) = (2.4e-11)(-9.8)

q = 1.6e-12

To find the number of electrons, we just divide by the electron charge, 1.602e-19.

Number of extra electrons = 1.6e-12/1.602e-19 = 9.9e6 = 1.0e7

51. An electron with speed v0 = 1.5 X10-6 m/s is traveling parallel to an electric field (v0 ][ E) of magnitude E = 7.7 X 103 N/C. (a) How far will it travel before it stops?  (b) How much time will elapse before it returns to its starting point?

a)

Given Info:

E = 7.7e3 N/C

q = 1.602e-19

m = 9.11e-31

F = ma = Eq

a = Eq/m

a = (7.7e3)(-1.602e-19)/(9.11e-31) = -1.35e15 ms-2

After finding a, we now have enough (3) variables to simply use SUVAT.

SUVAT

u  = 1.5e-6 m/s

v = 0

a = -1.35e15 ms-2

v2 = u2 + 2as

(0) = (1.5e-6)2 + 2(-1.35e15)(S)

8.3e-4 m = S

b)

To solve for time, again we use SUVAT.

u =  1.5e-6

v = 0

a = -1.35e15 ms-2

v = u +at

0 = 1.5e-6 + (-1.35e15)(t)

t = 1.11e-9

To find the time it takes to fall to its original position, we simply multiply by 2. (Similar to a rocket going up and down again... it takes twice the time of going up).

t = (1.11e-9)*(2) = 2.22e-9 (nanoseconds)