Four small spheres, each charged to +15 nC, form a square 2.0 cm ...

Question

Four small spheres, each charged to +15 nC, form a square 2.0 cm on each side. From far away, a proton is shot toward the square along a line perpendicular to the square and passing through its center. What minimum initial speed does the proton need to pass through the square of charges?

Accepted Answer

Hello, fellow physicists today, we're to solve the following practice problem together. So first off, let's read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. Three small spheres with charge of positive 10 nano Kums form a triangle with sides measuring 2.5 centimeters. An alpha particle helium nucleus is projected towards the triangle from a distance along a perpendicular line passing through the center of the triangle. What is the minimum initial speed necessary for the alpha particle to successfully traverse through the arrangement of charged spears? So that's our end goal is to find the minimum initial speed necessary for the alpha particle to successfully traverse through the center of the charged spears. Fantastic. So we're given some multiple choice answers and they're all in the same units of meters per second. So let's read them off to see what our final answer might be. A is 1.8 multiplied by 10 to the power of 12 B is 1.9 multiplied by 10 to the power of six C is 7.1 multiplied by 10 to the power of five and D is 4.9 multiplied by 10 to the power of eight. OK. So first off, let us write down all of our known and unknown variables. So charge is given to us as positive 10 nano Kums. And each side of the triangle is given to us as 2.5 centimeters. So I'm gonna call the size of the triangle lowercase S and that's equal to 2.5 centimeters. And that is the same as 0.025 m. You can use dimensional analysis to find this value or you can just like type it into Google really quick to find that. OK. So we do not know what the height of the triangle is and well, you know, it's draw a little triangle really fast. So, so we know that each of the sides is 2.5 centimeters. So that means that if all the sides are 2.5 centimeters, that means we're dealing with an equilateral triangle. So that means when that all the sides are the same and all the angles are the same. So that means that each angle has to be 60 degrees because 60 degrees multiplied by three is 180 degrees. And that's wonderful because all triangles add up to 180 degrees. OK. So at this stage, we need to recall the equation to help us to determine the height of a triangle. So let us remember that sign of 60 degrees since 60 since we're dealing with an e collateral triangle is equal to the height of the triangle which is lowercase H divided by 0.025 m. And to draw another little triangle that was visualized. So to find the height, we just draw a line, a vertical line to form a 90 degree angle to line up with the hypotenuse which is the side of the triangle, which is 2.5 centimeters and our ankle is 60 degrees. Awesome. So now we need to isolate and sulfur H. So when we rearrange and sulfur H we get H is equal to 0.025 m multiplied by sine of 60 degrees. So when we plug that into a calculator, we should get that the height is equal to 0.02165 m. So now we must consider the distance of the center of the triangle from any of the charges. So let us call this value R note that the center of the equilateral triangle will be the Centroid which will divide the median into a ratio of 2 to 1. Also note that for an equilateral triangle, the median will also be the height of the triangle. Therefore, we can write that R is equal to the height of the triangle divided by three. So plugging in our known variables. So 0.02165 m divided by three. When we plug into a calculator, we will get 0.00722 m. Now, at this stage, we can recall and apply the law of the conservation of energy. So that states that the initial kinetic energy, which I'm gonna denote kinetic energy as ke so the initial kinetic energy plus the initial potential energy, which I'm gonna denote as pe is equal to the final kinetic energy plus the final potential energy. OK. Which we can write as one half multiplied by the mass of the alpha particle, which I'm going to denote as M subscript, a multiplied by the initial velocity squared plus the charge Q plus Q multiplied by V I which is the initial velocity is equal to one half multiplied by ma multiplied by va VF squared. The final velocity squared plus Q multiplied by the final velocity. Now, if we assume that the initial potential is approximately zero. Since the alpha particle is fired from far away. As stated in the problem, we can go on to write that one half multiplied by ma multiplied by V I squared is equal to Q multiplied by VF Awesome. So now we need to rearrange this equation to solve for V I. So when we start to do that, we get that V I squared is equal to two multiplied by Q multiplied by VF all divided by ma. But we have to consider who's on it. And we have to consider the source charge which let me write the equation really quick. And then we can quickly explain a little in depth what the difference between lower case Q and capital Q is. But don't panic, we'll get there. I promise. So the initial velocity squared is equal to two multiplied by lower case Q multiplied by three multiplied by K which is Kim's constant multiplied by capital Q all divided by ma multiplied by R. OK. So let's pause here. So capital Q is the source charge. And lower case Q is the test charged used to measure the strength of the electric field at a given location surrounding the source charge. Awesome. And it's also we need to note that the final velocity VF is equal to Coolum constant multiplied by the source charge divided by R. And this is the electric potential at the center of the triangle due to charge queue. And since there are three charges, we have multiplied the equation by three. So that's where the three comes from. So we can then knowing all this information, we can rewrite the initial velocity squared is equal to two multiplied by two E where E is the elemental charge for a single proton multiplied by three multiplied by Kam's constant multiplied by the source charge. OK. So now to get rid of the squared symbol for the squared symbol on the initial velocity, all we have to do is just take the square root of the other side. So let's do that and solve for V I and plug in all of our known variables to solve. So two multiplied by two multiplied by three multiplied by the elemental charge which is 1.60 multiplied by 10 to the power of negative 19, negative 19 cool arms multiplied by Kums constant, which is 9.0 multiplied by 10 to the power of nine newtons multiplied by meters squared per Coolum squared multiplied by the charge, which is our charge given to us. In the problem was 10.10 multiplied by 10 to the power of negative nine coulombs means we had to convert nano columns to columns all divided by the mass. Oh, here. So we don't get confused. We forgot one important piece of information means it's the mass of the alpha particle multiplied by R Awesome. OK. So the mass of an alpha particle, when we look that up, we determine that to be 6.64 multiplied by 10 to the power of negative 27 kg multiplied by our R value which we determine to be 0.00722 m. Fantastic. OK. So when we plug in that big old Long equation into our calculator, our final answer will be 1.9 multiplied by 10 to the power of 6 m per second. His because our calculator did all that work. We got the right answer. Hooray. OK. So let's go look at our multiple choice answers to see what the final answer should be. So looking at our multiple choice answers, the correct answer has to be the letter B, 1.9 multiplied by 10 to the power of six meters per second. And if you're wondering where the seconds come from, it's what all the units in our problem. We just, the last problem we used will cancel out to give it to us beca will give us the meters per second mean we're solving for velocity. Awesome. Thank you so much for watching. Hopefully that helped and I can't wait to see you in the next video. Bye.

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Four small spheres, each charged to +15 nC, form a square 2.0 cm on each side. From far away, a proton is shot toward the square along a line perpendicular to the square and passing through its center. What minimum initial speed does the proton need to pass through the square of charges?

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