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. (a) Find the magnitude of the electric field.

Video transcript

Hi, everyone in this problem, we are asked to actually determine the magnitude of the electric field e between the two sheets where we have to infinite parallel sheets of opposite charge separated by 25 centimeters. And an electron is actually released from rest at the surface of the negative sheet and reaches the positive sheet three microseconds later. So we want to start with creating a list of everything that is given. So first, we have to distance which is 25 centimeter which is times 10 to the power of minus two m or at the same time, it can be 0.25 m. And then we will have an electron, an electron which is released from rest Sophie not X or zero X is going to Ashley equals to zero m per second and it will reach the positive sheet at three microseconds letter later. So we will have B equals three times 10 to the part of minus six seconds. And then that is everything that is given in the problem statement. So we then also know that an electron will have the mass Of 9.1 times 10 to the power of - kg just like so and then we can start solving this problem. So we have to recall that electrostatic forces or f equals QE are much stronger than gravitational forces. So the force of the gravity on the electrons can actually be neglected. So we want to apply Newton's second law. So on Newton's second law, we will have uh electrostatic forces plus the gravitational forces equals M times the total acceleration of a of the actual electron itself. So applying the Newton's second law, we will obtain the electric field in terms of M Q and A, especially because we can simplify this because we know that the electrostatic force is a lot bigger than the gravitational force. So we can essentially neglect this part. So F will equals to M A and F. The electrostatic force is Q multiplied by the electric field equals M A. So then we can rearrange this so that E will then equals to M A over Q just like. So next, we want to actually start including ah the acceleration into our kingdom attic equation because we are given the distance and also the time that it takes for the electron to reach from one side to the other or to go through the whole distance. So essentially, we want to employ the Kinnah Matic equation ask the following. So X minus X not or the total distance traveled equals half the acceleration times T squared plus fee, not T V nought X in this case, A X in this case, so this can definitely be A X as well. So we want to remember that the fee not X is zero. So we can neglect this part and simplify this even better. And then we want to take this in terms of A X. So rearranging this A X can definitely then B two times X minus X not or X zero over T squared just like. So And the next one, Ashley um substitute this A X into our electric field, E because that is what is being asked here. So A X can definitely be found by multiplying Q E over M. So we want to substitute this A X into this equation here to get what the E S in terms of our kingdom attic equation. So this is Q E over M equals two X minus X not over T squared. So we want to rearrange this in terms of E because that is what's being asked. So this is going to be two M X minus X not over Q T squared. So we know everything here because we are given all the information needed from the problem statement. So we can just immediately put it in. So X minus X not is essentially D. So I'm just gonna write that down here. So that will be 0.25 m. So two times M M is going to be the mass of the electron. So 9.1 times 10 to the power of minus 31 kg and then X minus X not is going to be 0.25 m or the distance over the Q Q is going to be the charge of electron. And in this case, we want to just use the magnitude. So this is going to be 1.6 times 10 to the power of minus 19. We want to only use the magnitude because we don't really care. We are only being asked the magnitude of the electric field and then the T is going to be three microseconds. So three times 10 to the power of minus six seconds squared, and that will be all for all the information and this will come up to an electric field value of 0.31 newton per cool. Um So We can conclude that the electric field between the two sheets is 0. Newton per Coolum. And they all correspond to option A that we have here. So that will be all for our particular problem. And if you guys have any sort of confusion, please make sure to check out our other similar lesson videos and that will be all for this particular video. Thank you.