Two 2.0-cm-diameter disks spaced 2.0 mm apart form a parallel-pla...

Question

Two 2.0-cm-diameter disks spaced 2.0 mm apart form a parallel-plate capacitor. The electric field between the disks is 5.0×10^5 V/m.b. An electron is launched from the negative plate. It strikes the positive plate at a speed of 2.0×10^7 m/s. What was the electron's speed as it left the negative plate?

Accepted Answer

Hello, fellow physicists today, we're gonna 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. A parallel plate capacitor is assembled using two square plates with a side length of 5.0 centimeters and a spacing of 1.0 millimeters between the plates. The electric field strength between the plates is 4.0 multiplied by 10 to the power of seven volts per meter. If a proton is shot from the negative plate towards the positive plate and arrives moving at 3.2 multiplied by 10 to the power of 6 m per second. What was the proton speed as it left the negative plate? OK. So we're given some multiple choice answers. 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 3.2 multiplied by 10 to the power of six B is 4.2 multiplied by 10 to the power of six C is 2.8 multiplied by 10 to the power of six D is zero and E is 1.0 multiplied by 10 to the power of six. OK. So first off, let us consider what exactly we're trying to solve for, for this particular problem. So this problem wants us to solve for the changes in speed. And this can be determined by using the relation to kinetic energy or momentum. However, for this particular problem, no collisions are taking place. So we cannot apply and use momentum conditions. However, we can apply and use the conservation of energy principle. Because kinetic energy and electric potential energy are related. Let us state that a charge Q is in a position where the electric potential is V and has electric potential energy of Q multiplied by VQV note that positive charges have positive electric potential energy and negative charges have negative potential electric potential energy. Making sure to indicate our sign on the chart, we can write that U is equal to positive Q multiplied by V and note that U becomes more positive as V increases and this is increasing towards a higher potential for positive charges. So we can then write that delta U is equal to U subscript F minus U subscript I. So delta U is equal to the final electric potential energy minus the initial potential energy. And this is more positive since the potential energy increases and kinetic energy decreases. So U becomes more negative as V increases and this is increasing towards a higher potential for negative charges. So delta U equal to UF minus U I is negative since the potential energy decreases and kinetic energy increases. So therefore, when we use the conservation of energy, we can write that the kinetic energy, the initial kinetic energy plus the initial potential energy is equal to the final kinetic energy plus the final potential energy. OK. So then we can write out, write this out even further, we can expand it. And we could write that one half multiplied by M multiplied by V I which is lowercase V I squared. So this is the initial velocity squared plus Q capital V I is equal to one half mm or one half multiplied by M multiplied by lowercase VF the final velocity squared plus Q multiplied by capital V F. OK. Where lower case V in this case, as I mentioned is the velocity and capital V is the electric potential. OK. So our end goal is to find the proton speed. So we need to solve for V I. So let's start working towards isolating and solving for V I by itself. So V I squared is equal to two divided by M multiplied by one half multiplied by M multiplied by lower case VF squared plus Q multiplied by capital V F minus Q multiplied by capital V I. And this is all equal to lowercase VF squared plus two multiplied by Q divided by M multiplied by capital BF minus capital V I, which is equal to VF squared plus two Q divided or two multiplied by Q divided by M multiplied by delta capital V. That was a mouthful. So now we need to note that delta capital V which to reiterate capital V is the electric potential. So since the electric potential, the delta value, delta capital V is unknown, we also we can write that the relationship between the electric potential difference. Delta capital V and electric field capital E and the plate separation lowercase D for the capacitor is so the equation for that. So putting into writing what we had just discussed out loud, so the electric field E is equal to delta capital V divided by the separation distance D. So now we need to rearrange this equation to solve for delta V. So delta capital V is equal to the electric field multiplied by the separation distance. So at this stage, we can plug in all of our known variables and solve. So let's do that. So delta capital V is equal to the electric field which that value is provided to us as 4.0 multiplied by 10 to the power of seven volts per meter. And this is in the problem itself multiplied by the separation distance which is 1.0 millimeters. But we need to quickly convert millimeters to meters. So let's perform dimensional ana analysis. So in 1 m, there is 1000 millimeters. So when we plug that into a calculator. We should get that delta capital V is equal to 40,000 volts. OK? And this is important. So I'm gonna box this value in blue. This isn't our final answer, but we need this to find our final answer. So this is important. So at this point, we need to find out what the sign is for delta capital V equals delta, delta capital V is equal to capital VF minus capital V I where capital VF is at the positive plate and has a high potential. And capital V I is at the negative plate and has a lower potential. Thus, we can write delta capital V is equal to capital VF minus capital V I is equal to a high. So it's a higher like a high potential minus a low potential. So we can write that this is equal to positive V multiplied by E which we can write as two multiplied by Q divided by M multiplied by delta capital V. And we need to note that Q and delta capital V are positive. Thus, the kinetic energy must increase. So then we can go on even further to write that V I or I should say lowercase V I, so that we're talking about the initial velocity again is equal to two divided by M multiplied by one half multiplied by M multiplied by lower case VF squared. The final velocity squared plus Q multiplied by the final velocity minus Q multiplied by the initial velocity. And this is all equal to the final velocity squared plus two multiplied by Q divided by M multiplied by the final velocity minus the initial velocity. And this is all equal to the final velocity squared plus two multiplied by Q divided by M all multiplied by delta capital V. OK. So now we can get rid of the squared sine on the initial velocity by square rooting the other sides. When we do that, we get that the initial velocity is equal to the square root of the final velocity squared plus two multiplied by Q divided by M multiplied by delta capital V. So now we can plug in all of our known variables and solve for V I. So let's do that the final. So we know our value for the final velocity but we need to find the initial velocity. So let's do that. So square root of, and then our final velocity was given to us in the promise 3.2 multiplied by 10 to the power of 6 m per second. And this is squared plus two multiplied by Q which our value for Q is 1.6 multiplied by 10 to the power of negative 19 columns divided by the mass of a proton. So the charge of a proton is 1.6 multiplied by 10 to the power of negative 19 columns. And then the mass of a proton is 1.67 multiplied by 10 to the power of negative 27 kilograms multiplied by our delta V value or delta capital V which we determine to be 40,000 volts. Awesome. So now when we plug in that into a calculator and solve, we should get 4.2 multiplied by 10 to the power of six meters per second. And that is our final answer. Let's box sat and green. So that was a lot of math. But if you break it down in the smaller terms, all make sense, I would just practice this from a couple times. It was all, it was a mouthful, but just think of it in terms of kinetic energy and potential energy and then it'll be a quick solve. Ok. So going back to the top to look at our multiple choice answers, the correct answer has to be the letter B 4.2 multiplied by 10 to the power of 6 m per second. 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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