Speed of electron in electric field question

1. Jun 27, 2012

lowcard2

1. The problem statement, all variables and given/known data
Two stationary positive point charges, charge 1 of magnitude 3.55 and charge 2 of magnitude 1.60 , are separated by a distance of 55.0cm. An electron is released from rest at the point midway between the two charges, and it moves along the line connecting the two charges.

What is the speed of the electron when it is 10.0 from charge 1?

The attempt at a solution

Fnet = kQe/r² - kqe/r² = [(9x10^9)(1.6x10^-19)/0.275²][3.8 - 2]x10^-9

Fnet = 3.71x10^-17

a = Fnet/m = 4.076x10^13

t² = 2ax = 2.855x10^6

t = (10^6)√6.32 = 2.855x10^6 m/s

v(x) = 2.855x10^6 m/s x 4.076x10^13

thank you all for your time

2. Jun 28, 2012

SammyS

Staff Emeritus
Hello lowcard2. Welcome to PF !

As the electron moves, the force on it changes, and therefore so does it's acceleration.

I suggest doing this problem using electric potential & conservation of energy.

3. Jun 28, 2012

lowcard2

thank you for the welcome!

forgot to include units in question. nC charges and cm distances
can someone check this?

Potential at a point: V = kq/r.

Relative to charge 1, the potentials are V1i = (9x10^9)(3.55)(10^-9)/(0.275) = 116 volts before, and
V1f = (9x10^9)(3.55)(10^-9)/(0.10) = 3.20x10^11319.5volts after.
Potential difference: 203.5V

Relative to charge 2, V2i = (9x10^9)(1.60)(10^-9)/(0.275) = 52.4 volts before, and
V2f = (9x10^9)(1.60)/(0.45) = 32 volts after.
Potential difference: -20.4V

Those differences are in opposite directions (one pulls left, the other right), so 203.5 -20.4V = 183.1V

Energy gained by an electron travelling down that gradient: qV = (1.6x10^-19)(183.1) = 2.93x10^-17 joules.

2.93x10^-17 = 1/2(9.11x10^-31kg)v^2
v=8.02x10^-6

4. Jun 28, 2012

SammyS

Staff Emeritus
How can you divide something on the order of 10-17 by something on the order of 10-34 and wind up with something much less than 1 ?

5. Jun 28, 2012

lowcard2

sorry thats not supposed to be ^-6. its positive

any input would be great. I only have one submission left so i have no room for error

Last edited: Jun 28, 2012
6. Jun 29, 2012

lowcard2

i have a couple hours to submit. any more comments? thank you guys