(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

We know that the magnitudes of the negative charge on the electron and the positive charge on the proton are equal. Suppose, however, that these magnitudes differ from each other by 0.00094%. With what force would two copper coins, placed 1.1 m apart, repel each other? Assume that each coin contains 3.2 × 10^22 copper atoms. (Hint: A neutral copper atom contains 29 protons and 29 electrons.)

2. Relevant equations

Coulomb's Law:

[itex]\textit{F = k}_{e}\frac{{q}_{1}{q}_{2}}{{r}^{2}}[/itex]

3. The attempt at a solution

Although my final answer was marked incorrect, here are the steps I took to get there. Hopefully someone will be able to help me spot it out :P

First, since the magnitudes of charges between protons and electrons differ by 0.00094%...

[itex]\frac{|q_{proton}|}{|q_{electron}|}\textit{ = 0.0000094}[/itex]

[itex]\textit{|q}_{proton}\textit{| = 0.0000094 × |q}_{electron}\textit{|}[/itex]

[itex]\textit{|q}_{proton}\textit{| = 0.0000094 × (1.6 × 10}^{-19)}\textit{ C}[/itex]

[itex]\textit{|q}_{proton}\textit{| = 1504 × 10}^{-27}\textit{ C}[/itex]

(Assuming that protons were the ones with the lower magnitude)

So then that means for one copper coin atom with 29 protons and 29 electrons, ...

[itex]\textit{n}_{p}\textit{ = Number of Protons}[/itex] & [itex]\textit{n}_{e}\textit{ = Number of Electrons}[/itex]

[itex]\sum{Q}\textit{ = (q}_{proton}\textit{ × n}_{p}\textit{) + (q}_{electron}\textit{ × n}_{e}\textit{)}[/itex]

[itex]\sum{Q}\textit{ = [(1504 × 10}^{-27}\textit{ C) × 29 protons] + [(-1.6 × 10}^{-19}\textit{ C) × 29 electrons]}[/itex]

[itex]\sum{Q}\textit{ = -4.64 × 10}^{-18}\textit{ C}[/itex] Which is the charge that one atom holds.

Multiplying, ...

[itex]\textit{Q}_{coin}\textit{ = (-4.64 × 10}^{-18}\textit{ C) × (3.2 × 10}^{22}\textit{ atoms) = -148,479 C}[/itex]

So one coin, under the conditions of the problem statement, will have a charge of -148,479 C?... This seems very high...

Going further with this...

[itex]\textit{F = k}_{e}\frac{{{Q}_{coin}}^{2}}{{r}^{2}}[/itex]

[itex]\textit{F = (8.99 × 10}^{9}\textit{ }\frac{{N}\cdot{m}^{2}}{{C}^{2}}\textit{) }\frac{{(-148,479 C)}^{2}}{{(1.1 m)}^{2}}[/itex]

[itex]\textit{F = 1.6 ×10}^{20}\textit{ N}[/itex]

Again, this was marked as incorrect by the HW website. Where did I go wrong? I'll need to get the next attempt right or I'll only be getting half credit for getting the units right :/

Any help or hints would be greatly appreciated!

Thank you!

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# Homework Help: Coulomb's Law - Electric Force Between Two Copper Coins

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