# Electric Potential related question

1. Dec 18, 2005

### kfan321

Well my plan was to use V = E d
So I need to find E and d and I'll have the problem solved.
I used V2^2 = V1^2 + 2ad
v2 is 0, v1 is 20.1, and i used 9.8 for a.
So d came out to 20.6. But I don't think that's correct because I didnt account for the acceleration due to the electric field? So how do I do that?
To solve for E, I wanted to use F/q but not enough info given. Thought about (KQ)/r^2 but not sure if this senario can be considered point charges...

2. Dec 18, 2005

### Gamma

Due to the electric field, now effective accelaration is changed. Use newton's second low to find a. (Hint: What are the two forces now acting on the object?)

3. Dec 18, 2005

### kfan321

Well you really haven't said anything I don't already know. Ive said force gravity and electric field will deccelerate the ball... and thats where Im stuck

4. Dec 18, 2005

### robphy

Did you draw a free-body diagram?
As you have determined, a is an unknown. d is also unknown. So, your single kinematic equation is not sufficient to find d. Note that you have not used the information about the time of flight. Is there another applicable kinematic equation that can be used here?

5. Dec 19, 2005

### kfan321

i've tried d = v1t + .5 at^2
still have 2 variables. So I did a system of equations with the first kinematic equation. Subbing d into a, hoping to find the new a and it came out to -9.8. lol...

...I'm all for not doing other peoples homework for them, but you guys are hiding the info like it's some top FBI confidential document.

6. Dec 19, 2005

### robphy

If you do things algebraically, you might find a single equation with one unknown. (That is, there may be a simpler starting point.)

In "Homework, Coursework, & Textbook Questions", you are nudged toward (but not given) the answer... but you have to show your work.

7. Dec 19, 2005

### kfan321

using the 2 kinematic equations previously

0 = v1^2 + 2*a*v1*t + a^2 * t^2

this is what i've done on post #5, i knew everything except a in that equation, so I solved for it and it came out to -9.8. Which can't be right b/c of the field, so where did I go wrong?

8. Dec 19, 2005

### robphy

Why is this true?

9. Dec 19, 2005

### kfan321

b/c at the top of the trajectory, it's 0 m/s, and I use time as 2.05 instead of 4.1...
But now I think about it, time shouldnt be half... Great, back to square one.

You know what, nvm the thread. I think Im better off figuring it out myself.