Three point charges are arranged along the x-axis. Charge q1 = +3.00 mC is at the origin, and charge q2 = -5.00 mC is at x = 0.200 m. Charge q3 = -8.00 mC. Where is q3 located if the net force on q1 is 7.00 N in the -x-direction?

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Hey, everyone today we're dealing with the problem about Cool Ems Law. So we're being told that in a virtual lab, small charge balls, which can be approximated two point charges are arranged as follows along the Y axis. Ball B one is placed at the origin and ball B two is at three centimeters. The charges of the balls are B one is to micro columns. B two is four negative four micro columns and B three negative six micro columns. The net force on the ball is also or on ball one is negative five newtons. With all this, we're being asked to find what the position of B three results in that net force. So before we go any further, one thing, we're going to be utilizing heavily his columns law which states that the force is equal to columns, constant multiplied by the absolute values of the magnitudes of the charges multiplied together, divided by R squared, the distance between the charges. So our values are given to us in micro columns. And we can recall that one micro column is equal to 10 Titty -6 columns. And with that, we can go ahead and list down our charges to the charge of Q one or ball one, two micro columns. So that will be two times 10 to the negative six columns. Q two, excuse me Is -4 times 10 to the negative 6th columns. Q three Is equal to - Times 10 to the -6 Columns. And the distance between R one or R or ball one and ball to which I'll write as R 21 Is negative three cm and recalled at one cm is equal to to the negative second meters. So this will simply be in or excuse me three times 10 to the 2nd m. And just for reference sake, I'll just draw out our little virtual lab, our plot. So we have the Y axis is what we're looking at. Let me just draw this here, the Y axis and we have our X axis for origin. So this is the origin point and less travel blue. So we'll say this, this right here is ball one in red. I'll draw ball too This ball to at three cm. And we now we need to find ball three and we can do this with the information given. So let's first find out what is the force acting upon ball one by ball too. Let's try this out. So force Ball to acting in ball one will simply be Colmes law. So we'll take nine times 10 to the ninth, which is the columns constant, we have native four or either or just fine. Let me do this. In our order, we have two times 10 to the negative six columns Multiplied by -4 times 10 to the AS- six columns divided by three times 10 to the 2nd m squared, Which will give us an answer of 80 newtons. So with that, we can go ahead and move forward. So if we're assuming we have a system at equilibrium, we can also say that the sum of all the forces in the Y direction because we're dealing with the Y axis will be ah The force of two on 1 in the wire direction plus the force of three on one in the Y direction. Again, this isn't, this is force. Let me clarify, this is force On one in the Y Direction on B one. So we know what F two or the force from two on one is we want to find out what the force on three on one is because the net force on ball one is listed as -5. Let me write that native five I didn't write. So therefore, If we rearrange now substituting in negative five newtons for The sum of the forces on ball one we get that F of three on 1, the Y direction is simply - Newtons -F 21 or sorry f of two on 1, excuse me, just five, Which is negative 5 -80 which gives us -85 Newtons. So what does this value tell us? Well, if the force from three on one is negative, then be three must be on the negative Y axis, it must be below so that it attracts the force that it exerts or so that the attractive force is negative, let's say, because that will drag it into the lower position. Oops. So we can go ahead and use columns law again and I'll scroll down a bit just so we have a little more room using columns law. Again, we can say that the force, let's write this in green because I've been using red and blue. So far, we can say the force of three on one, I'll leave out the Y for now, since everything we're doing is in the Y direction Would be cool is constant multiplied by Q three, Q one, excuse me, over the distance between we're three and one squared. However, we're trying to find the position of the, of the three that will result in this net force, which means we need to solve for R. So rearranging, we get that are three on one is equal to the square root of K into the magnitude of Q three into the magnitude of Q one divided by The force of three on 1. Simplifying this, we get the square root Of nine times 10 to the 9th. That's cool. It's constant uh charge of ball three is negative six times 10 to the native six or oops, we're dealing with magnitudes. So it will simply be six times 10 to the, The age of six. Also let me amend something before moving on during this first step when calculating the force of two on one, I wrote that it was a negative four, but we're dealing with magnitude. So it should simply be four times 10 to the negative six anyways back here, Back on the 4th of three on 1, we're trying to find the distance now And we know that the ball one has a charge of two times 10 to the Native six columns. We'll extend this a bit divided by the force which we calculated to be Need to 85 Newtons. So solving this, we get an answer of 3.56 cm and we know that B three has to be on the negative Y axis because it is attracting the forces negative, which means it is attracting ball one in the negative direction. So we can say that ball three, We can see the ball three, That ball three is located at negative 3.5, 6 cm or answer choice. D I hope this helps and I look forward to seeing you all And the next one.

Three point charges are arranged along the x-axis. Charge q1 = +3.00 mC is at the origin, and charge q2 = -5.00 mC is at x = 0.200 m. Charge q3 = -8.00 mC. Where is q3 located if the net force on q1 is 7.00 N in the -x-direction?

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Key Concepts

Coulomb's Law

Superposition Principle

Direction of Forces