Two small aluminum spheres, each having mass 0.0250 kg, are separated by 80.0 cm. (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 1.00 * 10^4 N (roughly 1 ton)? Assume that the spheres may be treated as point charges.

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Welcome back everybody. We are taking a look at two copper balls here and we are told a couple of different things, we're told that they are 45 cm apart or .45 m. And we are told that we wish to transfer electrons back and forth. So that an attractive force is established between them, meaning that they will have a positive charge and a negative charge on them. One thing we are also told however, is even though they do have different signs, the magnitude of charge A. K. The absolute value of each of the charges is going to be equal to The same thing equal to one another. Now, the attractive force that we want to feel is 8000 newtons and we are tasked with finding how many electrons should be transferred for this to occur. Luckily we have a formula for this that states that the magnitude of charge for just one of the walls is equal to the number of electrons that should be transferred times the magnitude of charge of an electron itself. So I'm actually going to divide both sides by E. Because we want to solve for N. And this gives us the N is equal to Q over E. But what is Q. While in order to find Q We are going to need to use Cool. Um law law states that the force is equal to a constant times the magnitude of charge of the first ball, times the magnitude of charge of the second ball, which is just Q Time Q or Q squared. This is all going to be divided by the distance between them squared. Now we need to isolate this Q terms. So I'm going to start by multiplying both sides by R squared over a. You'll see that on the right here, we can cancel out some terms and we are left with that Q squared is equal to F R squared over K two. Finally isolate Q all by itself. We need to take the square root of both sides and we get that our Q is equal to the square root of F R squared over K. So now we are ready to go ahead and plug in some numbers to find the magnitude of charge of one of the copper balls. So we have that Q is equal to the square root of our attractive force 8000 newtons times the distance between them which is 80000.45 squared all over columns constant of nine times 10 to the ninth. When you plug this into your calculator, you get that this is 4.24 times 10 to the negative fourth columns. Now we are ready to find the number of electrons that need to be transferred by this formula that we established earlier. So we have on top we have the magnitude of charge that we just found 4.24 times 10 to the negative fourth columns divided by the magnitude of charge of an electron, which is 1.6 times 10 to the negative 19th columns. Plugging this into our calculator. We get that this is equal to 2.65 times 10 to the 15th electrons, which corresponds to our answer choice of C. Thank you all so much for watching hope this video helped. We will see you all in the next one.

Verified Solution

Key Concepts

Coulomb's Law

Charge of an Electron

Point Charge Approximation