Some Electric Potential energy questions

1. Feb 13, 2006

wr1015

A charge of 3.19 µC is held fixed at the origin. A second charge of 3.19 µC is released from rest at the position (1.15 m, 0.530 m) .

(a) If the mass of the second charge is 3.10 g, what is its speed when it moves infinitely far from the origin?

ok so i've broken it up into x and y components and used pythagorean theorem to find the line that represents the force that the second charge exerts on the fixed charged $$\sqrt{1.15^2 + .530^2}$$ and my answer was 1.266 m. Then i used tan$$^-1$$ (.530/1.15) to find theta which came out to be 11.98$$^0$$. Then i calculated 1.266 (cos 11.98) for the positive x-direction and 1.266(sin 11.98) for the positive y-direction and i got 1.2384 m for the x and .26278 m for the y. Then i plugged those values back into pythagorean theorem again to find the net force $$\sqrt{.26278^2 + 1.2384^2}$$ and got 1.2666N. Then i used the Force per unit charge to find the net electric field and got (1.2666/3.19E-6) = 3.97E5 and multiplied by 2 since there are 2 charges and got 7.94E5 as my net electric field. And then i plugged that value into $$\sqrt{2q\Delta V/m}$$ to find velocity. Obviously i'm doing something wrong somewhere because i'm not getting the right answer... any help would be greatly appreciated.

Last edited: Feb 13, 2006
2. Feb 13, 2006

marlon

Hint : what is the potential if the charge is that far away from the first charge.

marlon

3. Feb 13, 2006

wr1015

well that's what i'm trying to figure out..

4. Feb 13, 2006

marlon

Well,

what's the forumula for the potential ?

Again, what's the right answer ? You told in your first post that you keep getting the wrong answer ? How do you know this ?

marlon

5. Feb 13, 2006

wr1015

formula for potential is $$\Delta U/q$$ but don't you have to figure out net electric field first since $$\Delta U = qEd$$ ??

i'm using webassign so the answers i keep entering are wrong

Last edited: Feb 13, 2006
6. Feb 13, 2006

marlon

The formula for the potential of a charge q felt at a distance r is $$U = k \frac{1}{r}$$ where k is a constant that does not matter here. Now, what's the U if q is at spatial infinity ?

It is important to understand the concept of potential (which is the aim of your exercise). A potential is actually potential energy per unit of electric charge. So two electric charges q1 and q2 have potential energy $$k \frac{q1*q2}{r}$$ where r is the distance between those two charges. If you devide this expression by, let's say, q1 you get the potential of q2. This potential, actually is a measure of how big the interaction (ie the magnitude of the potential energy) with an electrical charge will be if you put some charge q1 on a distance r from q2.

marlon

Last edited: Feb 13, 2006
7. Feb 13, 2006

wr1015

thank you for your help, i understand now..