CP A uniform electric field exists in the region between two oppo...

Question

CP A uniform electric field exists in the region between two oppositely charged plane parallel plates. A proton is released from rest at the surface of the positively charged plate and strikes the surface of the opposite plate, 1.60 cm distant from the first, in a time interval of 3.20 * 10^-6 s. (b) Find the speed of the proton when it strikes the negatively charged plate.

Accepted Answer

Hi, everyone in this problem, we are asked to find the speed of the particle one microseconds after being released where we have a tiny particle of mass five times 10 to the part of minus 15 kg and the charge of four nano column accelerated from rest between two opposite lee charged flat parallel sheets where there is a constant horizontal electric field of plus 2, +80 newton per cool. Um The two sheets are separated by a distance of five millimeter. And we are asked to find the speed of the particle. So the way we want to tackle this problem is to first by identifying that electrostatic forces, F equals Q E are much stronger than gravitational forces. So in this case, because the electrostatic forces are much stronger than the gravitational forces than the force of gravity on the tiny particle is going to be able to be neglected. So the way we want to solve for this problem is to buy first, just identify all the things that we know. So first we have the mass of the particle which is five times 10 to the power of minus 15 kg and a charge or a queue of four nano column. So four times 10 to the power of minus nine cool. Um Next, we will have an electric field of 2 80 Newton per column. So that will be the E 280 Newton per cool. Um It is a plus and two sheets are separated by a distance of five millimeter. So I'm just gonna write that down as A D five times 10 to the power of minus three m. And the T or the delta T essentially the T is going to be one microsecond one that's tended to part of minus six seconds. Okay. So the way you want to solve for this problem is to by recalling Newton's second law. So applying Newton's second law, we will obtain the acceleration in terms of M Q and E. So first, we have the electrostatic force plus the gravitational force MG which will equals to M multiplied by the acceleration M A. And as I said previously, the electrostatic forces F here which will equals two Q E is going to be much larger than the gravitational force. So I'm just gonna write this down Q E is going to be a lot larger than M G. So because of this, then we will be able to actually neglect M G. So the Newton's second law is essentially just going to be Q E equals to M A. And therefore the A can then be calculated by multiplying Q E dividing it by D M just like. So, so next to find the speed, which is the one that's being asked, we need to use the Kinnah Matic equation. And we want to first recall that the cinematic equation is going to equal to V X will equals two ft zero or V zero X plus A X times T. And this acceleration in terms of the X access will then be A X equals Q E over M. So we want to substitute that X into this equation here. So we will have three X equals fee, not fee zero, X plus Q E over M T. And in this case, we will have from, we will have the particle going from first arresting state or fee not X well equals to zero m per second. So because final X is zero, then this equation can then be simplified to be free X equals Q E over M times T. So we have all of this value from the information given. So we can actually calculate the fee X. So the Q is going to be four times 10 to the part of minus nine Coolum, The ES 280 Newton per Coolum and the M Is five times 10 to the power of -15 kg. And the T is the one microsecond, which is then to the part of minus six seconds. And therefore, we'll have three X to be 224 m per seconds. So that will essentially be the answer to our problem, which will correspond to option C just like. So, so option C will be the answer to our problem right now And that will be all. So the speed of the particle is going to be 224 m/s. And that will be all for this particular problem. If you guys have any sort of confusion, please make sure to check out our other lesson videos with similar topics and that will be all for this video. Thank you.

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