You have two small, 2.0 g balls that have been given equal but op...

Question

You have two small, 2.0 g balls that have been given equal but opposite charges, but you don't know the magnitude of the charge. To find out, you place the balls distance apart on a slippery horizontal surface, release them, and use a motion detector to measure the initial acceleration of one of the balls toward the other. After repeating this for several different separation distances, your data are as follows:Use an appropriate graph of the data to determine the magnitude of the charge.

Accepted Answer

Hey, everyone. Let's take a look at this practice problem. Two small objects each with a mass of 4 g and unknown charges are placed on a frictionless horizontal table at various distances apart. The magnitude of the charges on the objects is the same, but their signs are opposite. The resulting acceleration of one of the two objects is measured and recorded. The collected data is as follows. Then we're given a lovely little table showing off the distances in millimeters and accelerations in meters per second squared by analyzing the relationship between distance and acceleration using an appropriate graph, determine the magnitude of the charges on the objects. Option A four nano couls option B eight nano Coulombs, option C 12 nano Coulombs and option D 16 nano Coulombs. So the goal of this problem is to find the magnitude of the charges by coming up with some relationship between the distances apart they are and the accelerations that they experience from the force with the hint given with the hint given to us that a graph may be helpful. Now, first off, since we're dealing with forces and accelerations on electrically charged objects, the the big equation we need to recall for problems like this is Kam's Law, which states that the magnitude of an electrostatic force between two charged particles is equal to the cool and constant K multiplied by the product of the charges of the two particles. For our purposes, we know it's the same. So I'm just going to come up with one charge and then square it. And this is divided by the square of the distance between them, which I'm gonna call d, this is Kam's Law and it'll give us a relationship between the charge and the particles and the force between them. Another important formula to be aware of is Newton's second law which connects the acceleration into the equation. Recall that according to Newton's second law, the net force is equal to the mass of a particle multiplied by its acceleration. So that connects the acceleration to force. And since both the Kama's law and the Newton's second law are both equal to force, we can set both of these laws equal to each other and have one big relationship between all the variables that we're going to use in this problem. Now, according to the information given to us in the table, the acceleration of the charge changes based on the distance. So I'm gonna algebraically rewrite this formula for acceleration by dividing both sides of the equation by mass. So the acceleration is equal to the Coolum constant multiplied by the square of the charge. Q divided by the mass of the particle multiplied by the square of the distance between them. So now we have a nice neat little formula showing us how acceleration changes when the distance D changes. But we still need to know some other piece of information if we want to solve for a cue. And this is where drawing a graph begins to come in handy. If we take a look at our acceleration formula, you might notice that the form it's taking might look kind of familiar we have the output A equals some big quantity multiplied by a term that has the input in it, which is D if this looks familiar to you, it might be because this is very similar to the equation for a straight line where we'll often see the equation for a straight line in the form of output Y equals some term representing the slope multiplied by the input X and four straight lines. If we graph Y with respect to X, then whatever the slope of the line is, is going to be whatever the other quantity in between us. So if we want to solve for Q, what we can do is when we're making our graph is graph, all the different values for acceleration with respect to our input term one divided by D squared, find the slope of that graph and then set that slope equal to this middle term right here. And that is how we can solve for the electric charge. So the first thing we need to do is work out what our inputs will be. Our inputs will not our, our input will not just be the distance D it has to be one divided by D squared. Since that is the term that is standing in for the input X in our usual slope equation. So let's take all four of the distances that the problem has given to us and convert them from millimeters into meters so that our units are consistent and then also divide square them and then divide one by each of them. So for the distance of 25 millimeters doing the math, we find a distant or an inverted squared distance of 1600 inverse meters squared. For 50 millimeters, we find 400 over 75 millimeters, we find 178 and for 100 millimeters, we find 100. So now let's follow the problem's instructions and construct a graph going the acceleration on the Y axis and this inverted square distance on the X axis. So I'm setting up the axes, now this horizontal axis will be one divided by D squared with units of meters to the power of inverse square and the H then the vertical axis will be acceleration with units of meters per second squared. This origin point will be our zero. And since the accelerations will never go higher than one, I'm just going to set one to be the top of the axis, which means I'm just going to kind of split it up the way we tend to do. I put 0.5 there, a 0.75 there, make it a little more obvious. A 0.25 here, the horizontal axis is gonna be a little bit trickier, but since it ever goes higher than 1800 I'll set this to be 1800 and then kind of split it up accordingly. 1614 112 100 1000 800 604 100 and 200. Yeah, probably not perfect since this is just AAA crude sketch of mine, but this looks pretty decent, I think. Yeah. Now let's graph our four points to show this off. One of them is at a horizontal position of 1600 with an acceleration of 0.92 m per second squared. So it's going to be somewhere up here or just just about around this point right here. And if it helps something you can do is add little dotted lines connecting the point you're looking at to the axis that makes things easier usually does. So I recommend it. The next point has a horizontal position of 400 a vertical position of 0.23 m per second squared for the acceleration. Next point has a distance, a horizontal position of 100 78 at 0.1 m per second squared. So I'm gonna put this a little before the 200 point, then they be about half way up here. And the final point that 100 is about halfway up. And then to 0.05 at an acceleration somewhere around here is about right. So I'll fill in the points and to actually show off the line that we're taking. The slope of this is where we try to draw a line as straight as possible through the points. This is probably a lot easier if you were have on paper because then you can use a ruler. But yes, so this is our formula and this is our graph. Now, we want to find the slope of this line and we can do this by using the standard algebra one slope formula. So we pick any two Y positions preferably the ones farthest apart because that'll get you the greatest average over the data and then divide that by the corresponding difference in X values. It really doesn't matter too much though. So I'm going to do with the first two points. So we have accelerations of 0.92 meters per second squared minus 0.23 m per second squared. And we divide this by the corresponding difference in horizontal positions that is 1600 meters inverse squared minus 400. And if you put this into a calculator, we find a slope of about 5.75 multiplied by 10 rates. The power of negative 4 m cubed per second squared. So this is our slope. And as we discussed earlier, this is going to be equal to this entire quantity right here containing the Q term that we're trying to solve for in the first place. So to reiterate slope is equal to the Coolum constant times the multiplied by the charge squared divided by the mass M. So if we're trying to solve for Q, we just need to algebraically rewrite this to solve for the charge, we multiply both sides of the equation by M divide both sides by the Coolum constant K. And then take the square root of both sides to have an equation telling us that the magnitude of the charge is equal to the square root of slope multiplied by the mass N divided by the Coolum constant. So all this stuff for us to do is plug our values in. So as we just discovered the slope has a value of 5.75 multiplied by 10 to the power of negative four cubic meters per second squared. This is multiplied by the mass of the charge which is given to us as 4 g or alternatively, four multiplied by 10 to the power of negative 3 kg divided by the Coolum constant K, which has a value of 8.99 multiplied by 10 to the power of negative nine with units of Newton meters squared per Coolum squared. And if we put this into a calculator, we will find a charge of about 1.6 multiplied by 10 to the power of negative eight cools, which if we rewrite this using our unit prefixes, and this is equal to about 16 nano Coulombs. And this is our answer to the problem. And if you look at our multiple choice options, we can see that this agrees with option D. So option B is the correct answer to this problem. and that's all for this problem. I hope this video helped you out. If it did. Please consider checking out some of our other tutoring videos which will give you more experience with these types of problems. That's all for now. And I hope you all have a lovely day. Bye bye.

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