# How to find final kinetic energy for a charged particle

## Homework Statement

An alpha particle (helium nucleus, charge +2e) starts from rest and travels a distance of 2.5 cm under the influence of a uniform electric field of magnitude 12.5 kV/m. What is the final kinetic energy of the alpha particle?

## Homework Equations

U = 1/2 QV = 1/2 C(V^2) = (Q^2)/2C
Q = CV
C = (Eo A)/d

where U = potential energy, Q = charge, C= capacitance, A= area, d = distance, and V = potential difference

## The Attempt at a Solution

I tried using U = 1/2 QV by calculating that the charge is 2 times the constant charge of an electron (1.602x10^-19) and used 12.5 kv/m as V (but changed it back to just volts.
U = 0.5 (3.204x10^-19) (12500 v/m)
U = 2.0025x10^-15
But this is wrong. Also, in a practice version of the problem it said that given magnitude = 18.5 kv/m and distance = 1.6 cm the answer would be 9.48x10^-17.
Any help with this problem would be greatly appreciated.

lightgrav
Homework Helper
How can you get from the E-field (V/m) to an Energy (Joule = CV)?
This is not a Capacitor ... the general Potential Energy formula is U = Q_total V_average ...
(this gives a ½ for Energy STORED in a Capacitor, but at maximum charge Q and maximum voltage V.)