A parallel-plate capacitor with a 1.0 mm plate separation is c...

Question

A parallel-plate capacitor with a 1.0 mm plate separation is charged to 75 V. With what kinetic energy, in eV, must a proton be launched from the negative plate if it is just barely able to reach the positive plate?

Accepted Answer

Hello, fellow physicists today, we're gonna solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use. In order to solve this problem. A proton needs to travel from the negative plate to the positive plate of a parallel plate capacitor. The plates are separated by a 2.0 millimeter gap and are charged to 85 volts. What is the initial kinetic energy in units of electron volts that the proton must have in order to just barely reach the positive plate? So that's our end goal. Our end goal is we're ultimately trying to figure out what the initial kinetic energy is in units of electron volts that this specific proton has to have in order to just barely reach the positive plate. So f our final answer should be what the initial kinetic energy is for this specific proton. Awesome. So with that in mind, let's read off our multiple choice answers and let us note once again that they're all in the same units of electron volts. So A is 25 B is 33 C, is 42 and D is 85. OK. So first off, let us write down all of our known variables. So we know that the separation distance between the plates is D and that is equal to 2.0 millimeters, which is also equal to 2.0 multiplied by 10 to the power of negative three meters. So I've gone ahead and given you the conversion. But if you want to solver this yourself, you can use dimensional analysis or you can quickly look up the conversion yourself. But at the same time, I've just given you the conversion. Awesome. So now we can also note and recall that from the problem itself that the voltage across the capacitor and let's call this VV is equal to 85 volts. And we also know what the charge of a proton is. The charge of a proton, which we're gonna call Q is equal to 1.6 multiplied by 10 to the power of negative 19 coons. So now at this point, we need to recall and use the equation to calculate the electric field. So let us note and recall that the electric field capital E due to the charge on a capacitor can be written as E equals V divided by D. So now we can substitute in our known variables to solve for E. So when we do, we know that V is equal to 85 volts, we also know that D is equal to 2.0 multiplied by 10 to the power of negative three meters. So let me plug that into our calculator. We will get 42,000 500 volts per meter. Awesome. So now at this point, we need to recall and use the equation to help us calculate the kinetic energy, which we're gonna denote as ke of the proton, which we should recall this equation as ke the kinetic energy is equal to Q, the charge of the proton multiplied by the electric field multiplied by the separation distance. However, we need to recall and note that E equals V divided by D. Thus, we can simplify our equation to write that Ke is equal to Q multiplied by V. So now we need to plug in all of our known variables to solve for ke. So ke is equal to 1.6 multiplied by 10 to the power of negative 19 pums because that's our chart of a proton multiplied by 80 five volts because that's our value of B which when you plug that into our calculator, we will get 1.36 multiplied by 10 to the power of negative 17 jewels. But we have a bit of a problem here. Our final answer that we just solved for, for our kinetic energy is in units of Js. But we want our final answer in units of electron volts. So we need to recall and use dimensional analysis to help us convert jewels to electron bolts. So let's do that. So ke which is our kinetic energy. So ke is equal to 1.36 multiplied by 10 to the power of negative 17 jewels. And we need to recall that in 72 note that in order to convert jewels to electron volts, we need to divide our Juls value which is 1.36 multiplied by 10 to the power of negative 17 Juls. And we need to divide it by 1.6 multiplied by 10. So 1.6 multiplied by 10 to the power of negative 19 and its units are joules per electron volt. So that's our conversion factor. So as you can see, and I'll mark it in red. Our jewels units cancel out leaving us with just electron bolts, which is perfect. That's what we want our final answer to be in. So when we plug that into our calculator, we will get 85 electron volts as our final answer. And that's it we've solved for this problem. Hooray, we did it. So looking at our multiple choice answers, the correct answer has to be the letter D 85 electron volts. Thank you so much for watching. Hopefully that helped and I can't wait to see you in the next video. Bye.

A parallel-plate capacitor with a 1.0 mm plate separation is charged to 75 V. With what kinetic energy, in eV, must a proton be launched from the negative plate if it is just barely able to reach the positive plate?

Key Concepts

Electric Field in Capacitors

Work-Energy Principle

Kinetic Energy of a Charged Particle

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