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An electron has an initial velocity of $5.00\times 10^{6}\textrm{ m/s}$ in a uniform $2.00\times 10^{5}\textrm{ N/C}$ strength electric field. The field accelerates the electron in the direction opposite to its initial velocity. (a) What is the direction of the electric field? (b) How far does the electron travel before coming to rest? (c) How long does it take the electron to come to rest? (d) What is the electron's velocity when it returns to its starting point?
1. The electric field is in the same direction as the initial velocity of the electron.
2. $3.56\times 10^{-4}\textrm{ m}$
3. $1.42\times 10^{-10}\textrm{ s}$
4. $-5.00\times 10^{6}\textrm{ m/s}$
Solution Video

# OpenStax College Physics for AP® Courses Solution, Chapter 18, Problem 58 (Problems & Exercises) (5:16)

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This is College Physics Answers with Shaun Dychko. An electron is in an electric field and the electron has an initial velocity of 5.00 times 10 to the 6 meters per second and this field we are told accelerates the electron in the opposite direction to its initial velocity. So its acceleration is this way and the force therefore is that way as well and the force on an electron is in the opposite direction to the field direction because the field points in the direction of force that would be applied on a positive test charge but since the electron is negative, the field is going to be in the opposite direction to the force it applies on a negative charge. So the field must be to the right in order to have a force to the left on the electron the way I have drawn it here and so the field direction and the initial velocity direction are the same— that's the answer to part (a). And part (b) asks how far does the electron travel before coming to rest? So that means the final velocity is zero and we can use this kinematic equation to say the final velocity squared is the initial velocity squared plus 2 times acceleration times displacement and we can figure out acceleration using the net force is mass times acceleration and it's going to be the force applied by the electric field which is going to be negative q times E. Now when I'm substituting in for factors here, I am always going to be substituting in magnitudes so q is going to be positive 1.60 times 10 to the minus 19 electric field's gonna be positive as well. So in order to indicate the fact that this force is to the left, which is the negative direction according to our coordinate system where I have chosen positive to the right that means I have to introduce a negative sign here to say that. So we are going to divide both sides by m and we get then the acceleration is negative qE over m. So we can substitute that in for a in our kinematics formula— and we do that in red here— also writing in 0 for final velocity because it comes to rest and then we take all of this to the other side of the equation subtracting it from both sides, in other words. So we have negative 2qEd over m equals negative v initial squared and then do a bunch of work to solve for d by multiplying both sides by negative m over 2qE and we do that on both sides and then we get d is positive mv initial squared over 2qE. So then substitute in for each of these factors: the mass of an electron is 9.11 times 10 to the minus 31 kilograms times its initial velocity squared divided by 2 times the elementary charge times the electric field strength and that gives and that gives 3.56 times 10 to the minus 4 meters. And then in part (c), we want to know how long it takes the electron to come to rest? So we use a different kinematics formula that has the factor t in it. So we can subtract v initial from both sides here and then divide both sides by a afterwards and we get then the time is the final velocity minus the initial velocity divided by acceleration and the final velocity is zero so we substitute that in minus initial velocity and I just wrote this over 1 because I am going to multiply by the reciprocal of a instead of dividing by this fraction because I find dividing fractions by fractions confusing so I am instead going to multiply by the reciprocal of this fraction so multiplying by negative m over qE. So that's the substitution for a here and this works out to v initial times mass divided by elementary charge times electric field strength. So we substitute in for each of those factors and we get 1.42 times 10 to the minus 10 seconds. Then part (d) we are asked what is the electron's velocity when it returns to its starting point? So the final velocity squared is initial velocity squared plus 2 times acceleration times displacement but since it returns to its starting point, the displacement is zero so we have v f squared equals v i squared and take the square root of both sides and when you take the square root of both sides of an equation, it's actually v f equals the positive or the negative of v i— you can't say for sure— because either one of these squared will make v i squared positive. So if you were to take negative v initial and then square it that's gonna make v initial squared and if you took positive v initial and squared it that's also going to make v initial squared. So when you take the square root of v initial squared, you don't know what you started with; it could be this one, negative v i, or this one, positive v i, and so you get to use your knowledge of the physics of the situation to choose one of them and we are gonna choose negative because we know that this particle is gonna go to the right some distance and then return back and be going in the negative direction when it returns to its starting point. So v final equals negative v initial in other words.

Submitted by mililanicardenas on Sun, 10/11/2020 - 16:10

Why is the negative(-) charge of an electron not included in the final calculation? Charge of an electron = -1.602X10^-19

Submitted by ShaunDychko on Wed, 10/21/2020 - 06:58

Hello mililanicardenas,
Thanks for the question. At 1:26 I mention that I'm substituting magnitudes into the formula. The reason for this is that we want the negative sign for force to indicate direction. It's more strategic to consider the direction of the electric field, consider the sign of the charge, and use our understanding the direction of force on the charge due to the field to establish whether the force is negative (to the left in other words) or positive. This is just my personal approach, in that I usually prefer explicit negative signs in my formula, rather than expecting them to appear in the end after substituting positive or negative values. I prefer to substitute positive values usually, and put negative signs explicitly in the formula. But not always, so I can see why it's confusing, but I try to mention it when I'm doing this. If, at 1:26, you didn't add the negative sign to the formula as I did, but then substituted $-1.602\times 10^{-19}\textrm{ C}$, you would get the same correct answer.
All the best,
Shaun

In reply to by mililanicardenas