# Electrostatic Potential / Work Question?

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1. Jan 25, 2017

### lightofthemoon

1. The problem statement, all variables and given/known data
An electron and a proton are held on an x axis, with the electron at x = + 1.000 m and the proton at x = - 1.000 m. If a second electron is initially at 20 m on the x axis, and given an initial velocity of 350 m/s towards the origin, it does not reach it. How close to the origin does it come?

2. Relevant equations
U = k q1 q2 / r
W = q ∫ E dl
E = F / q
KE = .5mv^2

3. The attempt at a solution

I'm not quite sure how to approach this problem...

Calculate potential energy of the second electron:
U = k q1 q2 / r
U = (k * (1.6 * 10^-19) ^2 / 19) - (k * (1.6 * 10^-19) ^2 / 21)
U = 1.16*10^-30

Calculate initial kinetic energy
KE = .5mv^2
KE = .5 * 9.11 * 10 ^-31 * 350^2
KE = 5.58 * 10^-26

Calculate work needed to be done to bring it to the origin
W = q ∫ E dl
since all of this is done on x axis I think the equation will simplify to W = qEx ?
E = kq / r
E = (9 * 10^9 * 1.6 * 10^-19 / 21) - (9 * 10^9 * 1.6 * 10^-19 / 19)
E = -7.22 * 10 ^-12
W = 2.3* 10^-29

From here I'm not quite sure how to continue...

2. Jan 25, 2017

### lightofthemoon

I got the right answer through U / qE = x

But can someone explain why I didn't need to use the initial speed of the electron? (Or did I actually need it and just got the right answer by chance?) Thank you

3. Jan 26, 2017

### haruspex

It is not clear what you did to get the answer. In your original attempt you calculated the initial KE. Did you use that?

4. Jan 26, 2017

### lightofthemoon

No, I didn't use the initial KE at all.

Work to bring the particle to the origin = qEx
So I set U=qEx and solved for x (how far it would get to the origin)
x = U / qE
E = kq / r E = (9 * 10^9 * 1.6 * 10^-19 / 21) - (9 * 10^9 * 1.6 * 10^-19 / 19)
E = -7.22 * 10 ^-12
U = k q1 q2 / r
U = (k * (1.6 * 10^-19) ^2 / 19) - (k * (1.6 * 10^-19) ^2 / 21)
U = 1.16*10^-30

1.16*10^-30 / (-1.6 * 10^-19 * -7.22 * 10 ^-12 ) = x
x = 1.004 m