Coulomb's Law - Electric Force Between Two Copper Coins

[prosteve037]

Homework Statement

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.)

Homework Equations

Coulomb's Law:
\textit{F = k}_{e}\frac{{q}_{1}{q}_{2}}{{r}^{2}}

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%...

\frac{|q_{proton}|}{|q_{electron}|}\textit{ = 0.0000094}

\textit{|q}_{proton}\textit{| = 0.0000094 × |q}_{electron}\textit{|}

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

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

(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, ...

\textit{n}_{p}\textit{ = Number of Protons} & \textit{n}_{e}\textit{ = Number of Electrons}

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

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

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


Multiplying, ...

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


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...

\textit{F = k}_{e}\frac{{{Q}_{coin}}^{2}}{{r}^{2}}

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

\textit{F = 1.6 ×10}^{20}\textit{ N}


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!

 

[gneill]

The problem statement says that the charges differ from each other by the given percentage, not that one is that percentage of the other. So if p is the percent difference, it would be something like q_e - q_p = (p/100) q_e.

 

[prosteve037]

The problem statement says that the charges differ from each other by the given percentage, not that one is that percentage of the other. So if p is the percent difference, it would be something like q_e - q_p = (p/100) q_e.

Ohh okay. But is all the other math good?

Here's what I get if I use this in the same method:

\textit{q}_{p}\textit{ - q}_{e}\textit{ = 0.0000094}



Charge of Proton:

\textit{q}_{p}\textit{ = 940,000,000,000,016 × 10}^{-20}\textit{ C}



Net Charge of a Copper Coin Atom:

\textit{Q}_{net}\textit{ = (29) (940,000,000,000,016 × 10}^{-20}\textit{ C - 1.6 × 10}^{-19}\textit{ C) =}\frac{ 0.0002726 C}{atom}



Net Charge of a Copper Coin:

\textit{(0.0002726}\frac{ C}{atom}\textit{) × (3.2 × 10}^{22}\textit{ atoms) = 87,232 × 10}^{14}\textit{ C}



Force Between Two Copper Coins 1.1 m Apart:

\textit{F = (8.9875 × 10}^{9}\frac{{N}\cdot{m}^{2}}{{C}^{2}}\textit{) }\frac{{({87,232 × 10}^{14} C)}^{2}}{{(1.1 m)}^{2}}

\textbf{F = 5.65 × 10}^{47}\textbf{ N}

Does this sound reasonable? It's waaayy bigger than what my previous attempts were, by orders of magnitude.

Thanks

 

[Markus Hanke]

Did you consider the fact that protons and electrons have opposite charge ? It seems not...

 

[gneill]

Ohh okay. But is all the other math good?

The method looks okay after the charge difference is calculated. To be sure, you might consider doing the math symbolically first (just manipulate the variables) rather than plugging in numbers from the start. It's generally easier to spot algebra slips that way.

Here's what I get if I use this in the same method:

\textit{q}_{p}\textit{ - q}_{e}\textit{ = 0.0000094}

A percent difference is generally expressed as follows. Suppose you have two quantities A and B and you wish to ind the percent difference between B and A. The formula would be:

p = \left( \frac{A - B}{A} \right) \times 100

Note that the numerical difference (A - B) is divided by the original value, A. Multiplying by 100 makes it a percent (per hundred) so don't forget to take it into account.

 

[prosteve037]

Did you consider the fact that protons and electrons have opposite charge ? It seems not...

I thought about this and wasn't sure. Would the sign of charge matter from the very beginning? I didn't consider it when calculating the percent difference but I did consider it in finding the net charge of each copper coin atom.

The method looks okay after the charge difference is calculated. To be sure, you might consider doing the math symbolically first (just manipulate the variables) rather than plugging in numbers from the start. It's generally easier to spot algebra slips that way.

A percent difference is generally expressed as follows. Suppose you have two quantities A and B and you wish to ind the percent difference between B and A. The formula would be:

p = \left( \frac{A - B}{A} \right) \times 100

Note that the numerical difference (A - B) is divided by the original value, A. Multiplying by 100 makes it a percent (per hundred) so don't forget to take it into account.

Ah, I forgot to account for the denominator \textit{q}_{e}.

...

I get 1.4 × 10¹⁰ N with the denominator accounted for. I've already tried that answer though, and it was wrong :[

 

[gneill]

...

I get 1.4 × 10¹⁰ N with the denominator accounted for. I've already tried that answer though, and it was wrong :[

It's possible that the software is being picky about the significant figures and accuracy in the final digit. When I calculate the value in the same way I obtain 1.451 x 10¹⁰N, which might be rounded to 1.5 x 10¹⁰N for 2 significant figures.

 

[prosteve037]

It's possible that the software is being picky about the significant figures and accuracy in the final digit. When I calculate the value in the same way I obtain 1.451 x 10¹⁰N, which might be rounded to 1.5 x 10¹⁰N for 2 significant figures.

Wow, this actually was the problem haha, finally got it. How stupid :P

Thanks so much!