# Charged particle interaction

## Homework Statement

What is the force of attraction between a uranium nucleus(atomic number 92)
and an electron at a distance of 0.1 nm?

F = kQ1Q2/r2

## The Attempt at a Solution

F= 8.99*109*92*1.6*10-19*(-1.6*10-19)/0.1*10-9

Please let me know the attempt at a solution is correct.

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well, using Coulomb's Force(like you did), you get the equation

$$F=\frac{1}{4 \pi \epsilon_{0}} \cdot \frac{Q_{U}Q_{e^{-}}}{r^{2}}$$

Which then, by substitution:

$$F=\frac{1}{4 \pi \times 8.85 \times 10^{-12}} \cdot \frac{(92 \times 1.6 \times 10^{-19}) \cdot (-1.6 \times 10^{-19})}{(0.1 \times 10^{-9})^{2}}$$

Yeah, it looks like you just forgot to square the radial distance.