Why is my book using different equations to solve for velocity?

1. Sep 13, 2013

outxbreak

I am having a difficult time trying to figure out why my book says that I should solve for v by:
(1) Set the kinetic energy equal to the change in potential energy and solve for v

Then 3 problems later it says solve for v by:
(2)Set ΔK = −ΔU and solve for v

http://img593.imageshack.us/img593/1995/1idz.jpg [Broken]

I just need to know why they are using different equations to solve for v.. this is not making any sense.

Last edited by a moderator: May 6, 2017
2. Sep 13, 2013

Andrew Mason

They are essentially the same equations. ΔK = −ΔU means that the gain in kinetic energy is equal in magnitude to the decrease in potential energy.

The V (upper case) in problem 21 is the potential difference. The v (lower case) is the velocity of the charged particle.

AM

3. Sep 14, 2013

outxbreak

But how do I know when to use
ΔK = −ΔU
or
ΔK = ΔU

That is my problem. :'(

4. Sep 14, 2013

vela

Staff Emeritus
The book's being sloppy in the first case. It's really using $\Delta K = \lvert \Delta U \rvert$. For both the electron and proton, the potential energy decreases, so $\Delta U < 0$.

The proton rolls down electric potential hills, so Vfinal is less than Vinitial. ΔV is therefore negative, and ΔU = e ΔV is also negative. The electron rolls up electric potential hills, so Vfinal is greater than Vinitial. ΔV is therefore positive. The electron has a negative charges, so ΔU = -e ΔV again is negative. Because ΔU is negative, if you try to use $\frac{1}{2}mv^2 = \Delta U$, there'd be no real solution for v. You need to use $\Delta K = \frac{1}{2}mv^2 = -\Delta U$. In other words, you always use $\Delta K = -\Delta U$.

In electromagnetics, it's pretty easy to make sign mistakes if you try to rely solely on the math. It's often easier simply to neglect the signs and calculate the magnitude of quantities and then insert the correct sign based on physical intuition. For instance, in both cases here, we know that the particles are going to speed up, so their kinetic energy increases. This increase comes at the expense of a decrease in potential energy. To find the speed, we don't really care about the sign of ΔU; we just need its magnitude.