Electric Potential Energy Question

[Alouette]

Homework Statement

A uranium nucleus in a reactor captures a slow neutron and divides, or fissions, into two smaller daughter nuclei. Assuming the nucleus divides into two equal daughters with charge Q=46e and diameter d=2x10-14m, calculate their electric potential energy.

Homework Equations

k = (9x10^9)Nm^2/C^2

V = kQ/r

U = qV

The Attempt at a Solution

I've tried using the two equations above to find the U, but it's not right with both:

k(46)/(1x10^-14) & k(46^2)/(1x10^-14).

Am I using the wrong formulas? Not doing enough work?

 

[vivekrai]

May be you are using wrong units? What is the unit of charge?

 

[Alouette]

For the charge it just says Q=46e. So I just plug that into the equation:

U = (k*Q)/(r)

= (9x10^9)*(46^2)/(1x10^-4)

//Using the constant k, two charges of Q since there are two daughters, and dividing the diameter by 2 to get the radius.

Given the formulas they gave us in class, I would think they would be relevant to the question... this is why physics frustrates me! :(

 

[vivekrai]

e = Charge of an electron = -1.6 * 10 ^-19 C

 

[ehild]

In Q=46e e means the elementary charge, 1.6x10^-19 C. Are you sure that the formula for U is valid in this case? That charge 46e means 46 protons in both nucleus.
ehild

 

[Alouette]

Ah, stupid mistake again. Clearly these questions are clouding my basic understanding...

So let me try to understand, Q=46e = 46(1.6x10$^{-19}$) ?

And I have this formula too:

U= k*Q$_{1}$*Q$_{2}$*e$^{2}$/d$^{2}$

(Using d$^{2}$ since it should be both radius added together anyways)

So therefore:

U = (9x10$^{9}$)*46*46*e$^{2}$/(2x10$^{-14}$)

?

Yes seems so! Thanks again guys, I really appreciate your time for helping me understand. I'm trying my hardest to learn.