For All Your Engineering Needs

Here are the mastering physics solutions for this assignment on point charges…

Forces in a Three Charge System:

Part A:

What is the net force exerted by these two charges on a third charge q_3 = 52.0 nC placed between q_1 and q_2 at x = -1.235 m?

Your answer may be positive or negative, depending on the direction of the force.

Explanation and Solution:

This is a fairly basic question and is similar to the problems on the homework and in recitation. You will be using your most basic formula for the force due to charge  and simply adding the two forces that you find together to get the answer. You can see this in my work:

F = k (Q_1*Q_2) / d^2

We will start with our first pair of charges. It is important to remember what we are looking for throughout solving this problem problem. We want to know the force on q_3 due to q_1 and q_2 so don’t go mixing up which charges to use. In each formula you will need q_3 otherwise you aren’t finding what you want. We will solve for the first force as follows:

F = k (Q_3*Q_1) / d^2

F = k (52nC * 10nC) / .425^2

F = – 2.59 * 10^-5 N

Now we will solve for the second force on q_3:

F = k (Q_3*Q_2) / d^2

F = k (52nC * 32nC) / 1.235^2

F = – 9.82*10^-6 N

Now we are going to sum the two forces to find the net force on the charged particle:

F_1+F_2 = F

• 2.59 * 10^-5 N + – 9.82*10^-6 N = – 3.57*10^-5 N

Electric Field Conceptual Question:

Part A:

For the charge distribution provided, indicate the region (A to E) along the horizontal axis where a point exists at which the net electric field is zero.

Explanation and Solution:

As you can see by the diagram, there are two positive charges on the x-axis  and you need to know where the net electric field is zero. So since they are equal charges the point where the electric field is zero must be at a place where the distance from each of the charges is equal. From this we can determine that the point where the net electric field is zero is at point C.

Part B:

For the charge distribution provided, indicate the region (A to E) along the horizontal axis where a point exists at which the net electric field is zero.

Explanation and Solution:

Now we have two charges on the x-axis, however, one of them is +2q and the other is only +q. So the +2q charge will have a stronger electric field and push the particle more more than the +q charge so the zero point will be to the left of C but still between the +q and the middle point.

Part C:

Explanation and Solution:

For the charge distribution provided, indicate the region (A to E) along the horizontal axis where a point exists at which the net electric field is zero.

Explanation and Solution:

Nowhere.

Part D:

For the charge distribution provided, indicate the region (A to E) along the horizontal axis where a point exists at which the net electric field is zero.

Explanation and Solution:

A

Charge Aluminum Spheres:

Two small aluminum spheres, each of mass 0.0250 kilograms, are separated by 80.0 centimeters.

Part A:

How many electrons does each sphere contain? (The atomic mass of aluminum is 26.982 grams per mole, and its atomic number is 13.)

Explanation and Solution:

This is a fairly simple problem that just requires some thinking. Basically you want to find how many molecule of aluminum there are since you know how many electrons are in an aluminum molecule you can find the total amount in the sphere. Start by finding the amount of moles then multiplying by Avagadro’s Number then by the atomic number of aluminum:

25 g / (26.982 g/mol) = .9265 moles

.9265 moles * N_A =  5.58*10^23 molecules

5.58*10^23 * 13 = 7.25*10^24 electrons

Part B:

How many electrons would have to be removed from one sphere and added to the other to cause an attractive force between the spheres of magnitude 10000 N (roughly one ton)? Assume that the spheres may be treated as point charges.

Explanation and Solution:

This is just like most of the problems in this assignment but with one extra step. We are going to use the formula we have been using throughout all of these. Knowing that the force between these two objects has to be 10,000 N and the distance between them is .8 m we can solve for the charge difference necessary to produce that kind of force by assuming they have the same original charge.

F = K (Q_1^2) / d^2

10,000 N = K (Q_1^2) / .8^2

Q_1 = 8.43 * 10^-4 C

Now all we need to do is divide number by the charge of one electron to find the number of electrons it would take to produce this kind of charge:

N_e = (8.43 * 10^-4) / (1.60 * 10^-19)

N_e = 5.27 *10^15 electrons

Part C:

What fraction of all the electrons in one of the spheres does this represent?

Explanation and Solution:

This is a really easy question and its basically self-explanatory, you just have to divide your answer to the second one by the answer you got for the first one.

(5.27 * 10^15) / (7.25*10^24) =  7.26 * 10^-10

Conceptual Question 26.11:

Metal sphere A in the figure has 4 units of negative charge and metal sphere B has 2 units of positive charge. The two spheres are brought into contact.

Part A:

What is the final charge state of each sphere?

Explanation and Solution:

When you bring two items of different charges into contact like this problem its a lot like the thermal equilibrium problems we have solved in the past. Essentially they are each going to arrive at a final charge. Now we just have to figure out how to determine what charge that is. I have a method for solving this, we first total the charges (a + gets a +1 and a – get -1) doing this we find that we get a charge of -2. Now since we have two particles we assume the charge is evenly distributed among them so -2 divided by 2 is -1. Each particle has a charge of -1.

Part B:

(see above)

Problem 26.5:

Part A:

What is the total charge of all the protons in 2.6 mol of O_2 gas?

Explanation and Solution:

This problem is similar to the aluminum one. We just want to find the total number of O_2 molecules and multiply it by the number of protons in each molecule, don’t forget its O_2 not just O. Then we multiply that by the charge of one proton:

2.6 mol * N_A = 15.66 * 10^23 molecules

15.66 * 10^23 * 16 = 25.1 * 10 ^24 protons

(25.1 * 10^24) * (1.6 * 10^-19) = 4 * 10^6 C

Problem 26.14:

Two small plastic spheres each have a mass of 1.6 g and a charge of -50.0 nC. They are placed 2.2 cm apart (center to center).

Part A:

What is the magnitude of the electric force on each sphere?

Explanation and Solution:

This is a fairly simple problem that goes straight to the basics of this chapter. We will simply use our most basic equation as follows:

F = K(Q_1*Q_2) / d^2

F = K(50nC * 50nC) / (.022)^2

F = .0465 N

Part B:

By what factor is the electric force on a sphere larger than its weight?

Explanation and Solution:

They pretty much tell you how to do this question in the answer space. You just need to do F/mg:

= (.0465) / (.0016 * 9.81)

= 2.96

Problem 26.48:

The net force on the 1.0 nC in the figure charge is zero.

Part A:

What is q?

Explanation and Solution:

This one is a bit more complex than the last. We know that the net force on the 1 nC charge is zero and we are given two other charges and there distances from the other charges. We are asked to find the charge of q that will result in this zero net force on the 1 nC charge. Lets first think about this like vectors which is essentially what these are just with a different equation for force. Since the two 2nC charges are symmetrical we can ignore the x-components of there force vectors since they will just cancel, so now we know that the sum of the forces in the y direction must be zero and we know the charges and distances of the 2 nC charges as well as the distance of the unknown charge  so we should be able to solve for the charge:

F_1 = K (2nC * 1nC) / .0013

F_1 = 1.38 * 10^-5 N

Now we need to find the y-component of the force, so using trig:

tan^-1(2/3) = 33.7

F_1_y = F_1(sin33.7)

F_1_y = 7.68 * 10^-6 N

Using this force we set up the following equation to solve for the charge q:

2(F_1_y) + F_q = 0

1.53 * 10^-5 N + K(1nC * Q) / .02^2 = 0

Q = .68 nC

Problem 26.25:

Part A:

What is the strength of an electric field that will balance the weight of a 2.9 g plastic sphere that has been charged to -4.6 nC?

Explanation and Solution:

So we are asked to find how strong an electric field has to be to balance a weight of a 2.9g ball. Lets start by looking at the equation for the strength of an electric field:

E = F / q

We can see from this formula that the strength of an electric field is equal to the force over the charge. So what is our force in this case? Well our force is just mg, the weight of the plastic ball. And our charge is given to us in the problem statement as 4.6 nC. (I don’t explain this and I probably should have but the equations we use always have an absolute value around the charge, this is why even though its a negative charge I use a positive one) Plugging these values into the equation shown above we get the following:

E = (.0029 * 9.81) / (4.6 nC)

E = 6.2 * 10^6

Part B:

What is the direction of an electric field that will balance the weight of a 2.9 g

plastic sphere that has been charged to -4.6 nC?

Explanation and Solution:

Because the force to balance the weight of ball must be directed upward and the charge is negative, the electric field at the location of the plastic ball must be pointing downward. This is what the professor explained in class the other day about how to see where the electric field is “pointing” depending on the charge of the field. Hopefully you listening and not talking like the dicks in the back.