Change the chapter
Question
Two point charges $q_1$ and $q_2$ are 3.00 m apart, and their total charge is $20\textrm{ }\mu\textrm{C}$ . (a) If the force of repulsion between them is 0.075N, what are magnitudes of the two charges? (b) If one charge attracts the other with a force of 0.525N, what are the magnitudes of the two charges? Note that you may need to solve a quadratic equation to reach your answer.
Question by OpenStax is licensed under CC BY 4.0.
  1. One charge is $15.0\textrm{ }\mu\textrm{C}$ and the other is $5.01\textrm{ }\mu\textrm{C}$
  2. One charge is $35.0\textrm{ }\mu\textrm{C}$ and the other is $-15.0\textrm{ }\mu\textrm{C}$
Solution Video

OpenStax College Physics for AP® Courses Solution, Chapter 18, Problem 40 (Problems & Exercises) (8:20)

Sign up to view this solution video!

Rating

No votes have been submitted yet.

Calculator Screenshots

OpenStax College Physics, Chapter 18, Problem 40 (PE) calculator screenshot 1
OpenStax College Physics, Chapter 18, Problem 40 (PE) calculator screenshot 2
OpenStax College Physics, Chapter 18, Problem 40 (PE) calculator screenshot 3
OpenStax College Physics, Chapter 18, Problem 40 (PE) calculator screenshot 4
Video Transcript
This is College Physics Answers with Shaun Dychko. Two point charges are 3.00 meters apart and when you add the charges together, their sum is 20 microcoulombs and we have to figure out what each of the charges are given that there's a force of repulsion between them of 0.075 newtons. So Coulomb's Law says that the force or the magnitude of the force between two charges is Coulomb's constant times the absolute value of the charges multiplied together divided by the distance between them squared. Now when you are solving an absolute value equation, you have to consider two different scenarios: you don't know whether the part in between the absolute value signs is positive or if it's negative and so you need to consider both scenarios but we are gonna consider just the positive one here and we'll see that it works out nicely and I was taking positive because I drew this picture and took positive to the right, put down the two charges here and showed that the force when it's positive is the repelling force. So... yeah! I mean I could have put the force over here of the same magnitude and then I would have said that well the force is negative but this is the central issue with an absolute value equation— it can be either positive or negative, you don't really know— but we'll see that assuming that it's positive gives us the correct charges and had you chosen negative, you would have found that the charges in your answer would be opposites whereas when you take positive for the force, we'll find that the charges are both positive which we know is necessary in order for the force to be repulsion. Okay! So q 2 is 20 microcoulombs minus q 1 rearranging this up here and writing times 10 to the minus 6 in place of the prefix 'micro' and then we can substitute for q 2 and write it as 20 times 10 to the minus 6 coulombs minus q 1. So now we have an equation with only one unknown which is q 1. We'll multiply both sides by r squared and then distribute this factor into the brackets and so we have k times q 1 times 20 times 10 to the minus 6 coulombs minus k times q 1 squared. And then we will take this to the right hand side or subtract it from both sides and then switch the sides around and... (oh sorry that's not what I'm doing) what I'm doing is I'm taking this to the left hand side to make it positive because it's nice having a leading coefficient for the squared term to be positive and then take this to the left side as well which makes it negative and then also dividing everything by the Coulomb's constant, k. So we are left with q 1 squared there's no k there now because we divided it by k minus q 1 times 20 times 10 to the minus 6 coulombs— no k again because we divided it by k— and then this term also gets divided by k which we have to just write there and so we have plus r squared times force divided by Coulomb's constant all this equals 0 and then we can plug in numbers. And we have 3.00 meters—distance between the charges— square that times 0.075 newtons of force divided by Coulomb's constant, this works out to 7.510 times 10 to the minus 11 coulombs squared and this is a quadratic equation where the variable is squared and then there's a variable to the power of 1 and there's a constant term in which the variable does not appear and we can solve this using the quadratic formula. So the traditional letters to use for the quadratic formula is x equals negative b plus or minus square root b squared minus 4ac over 2a where b is the coefficient of the linear term which is the term to the power of 1. In this case, b is negative 20 times 10 to the minus 6— negative because of this minus sign there— c is positive 7.510 times 10 to the minus 11 and a is 1. So when we plug these numbers into this quadratic formula, we find that q 1 is 15.0 microcoulombs or this other charge here... so one charge is 15.0 microcoulombs— we don't know whether that's q 1 or q 2— and the other charge is 5.01 microcoulombs. Okay! And these add up to 20 as we expected and they are the same sign— they are both positive and so their force is repulsion— and so we can be confident that this is the correct answer. Part (b) says that the charges attract each other so based on this picture, we should take the force of attraction to be negative because it's in the negative direction although like I said before, the picture itself is not a reason to say the force is negative. I mean because we could have drawn the force over here in which case it would be positive but you have to try one and if that doesn't work out, you try the other until you get an answer where the charges are of opposite sign and that will happen based on experience of doing problems like this; I know that will happen when we choose negative for the force. So same sort of process as we had before: we write down the force and it's negative kq 1q 2 over r squared; we need to put the negative explicitly in our formula because q 1 and q 2 are magnitudes, you don't normally plug in the negative... so many negative coulombs into either these charges here. Okay so because those are magnitudes and we want the force to be negative, we need to introduce this negative sign ourselves which is okay because that's what you do when you solve an absolute value equation is you solve it once for having it positive and you solve it again for having it negative. Okay and so this scenario is solving it for assuming the product q 1q 2 is negative. Let's plug in 20.0 times 10 to the minus 6 coulombs minus q 1 in place of q 2 and then multiply both sides by r squared and we get this line and then this leading coefficient is already positive so let's bring this to the right hand side subtracting it from both sides in other words and then divide everything by k and you end up with this line here after switching the sides around. So we have q 1 squared minus 20.0 times 10 to the minus 6 coulombs times q 1 minus r squared F over k. So there's two differences in this part (b) versus part (a); there's now a minus here where there was a plus before and the force magnitude is different, it's 0.525 newtons now instead of 0.075. So then we work out what this is: it's negative 5.257 times 10 to the minus 10 plug the coefficients and this constant term into our quadratic formula and so we have 20.0 times 10 to the minus 6 coulombs plus or minus square root of 20.0 times 10 to the minus 6 squared minus 4 times negative the constant term c is a minus here because of this minus and then divide all that by 2 and one of the charges is 35.0 microcoulombs and the other charge is negative 15.0 microcoulombs. And a way to do a reality check on this is to ask: do they add up to 20? And the answer is yes because 35.0 plus negative 15.0 is 20.0; and will they attract? And the answer to that is yes because they have opposite signs.