# Challence problem on electric potential

• mrjoe2
Ke^2e/r)In summary, two particles, a proton and an alpha particle, with charges of +2e and masses of 4u, respectively, are fired towards each other from a far distance with an initial speed of 0.010c. The distance of closest approach between the two particles, measured between their centers, can be calculated by equating the total kinetic energy of the system to the potential energy at closest approach. This results in a distance of 1.93 * 10^-14, which differs from the textbook's answer of 1.93 * 10^-14.

#### mrjoe2

Edit to the title: CHALLENGE problem on electric potential energy*

## Homework Statement

A proton and an alpha particle (q = +2e, m = 4u) are fired directly toward each other from far away, each with an initial speed of 0.010c. what is their distance of closes approach, as measured between their centers?

## Homework Equations

u = proton mass = 1.67262158 × 10-27 kilograms
e = elementary charge = 1.60217646 × 10-19 coulombs
Kinetic energy Ke= 1/2 mv^2
U = Kq1q2/r
K = 9.0exp9
deltaKe + deltaU = 0

## The Attempt at a Solution

(deltaKe1 + deltaU1) + (deltaKe2 + deltaU2) = 0 where 1 represents the u mass proton and 2 represents the alpha particle.
((0-.5u(.01c)^2) + (Ke*2e/r)) + ((0-2u(0.01c)^2) + (Ke^2e/r)) = 0
solve for r =2.45 *10^-14

the answer in the textbook is 1.93 * 10 ^-14

Hi mrjoe2!
mrjoe2 said:
A proton and an alpha particle (q = +2e, m = 4u) are fired directly toward each other from far away, each with an initial speed of 0.010c. what is their distance of closes approach, as measured between their centers?

((0-.5u(.01c)^2) + (Ke*2e/r)) + ((0-2u(0.01c)^2) + (Ke^2e/r)) = 0

ah … but they're not stationary at closest approach, are they?

If you consider a proton and an alpha particle as a system, at the closest approach the total kinetic energy of the system is converted to the potential energy.
Hence
.5u(.01c)^2) + + 2u(0.01c)^2) = (Ke*2e/r))