How Many Electrons Calculate a Charge of -1.00 x 10^16 C?

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SUMMARY

The discussion centers on calculating the number of excess electrons on two identical water drops, each with a charge of -1.00 x 1016 C. The formula used is n = q / -e, where e is the elementary charge, with values of 1.602 x 10-19 C and 1.60 x 10-19 C being debated. The final answer is determined to be 625 electrons, aligning with the book's answer. The conversation emphasizes the importance of significant figures in calculations, suggesting that the constant's precision should match that of the given data.

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ChiralWaltz
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Homework Statement



Two tiny, spherical water drops, with identical charges of -1.00 x 1016 C, have a center-to-center separation of 1.00 cm.(a) What is the magnitude of the electrostatic force acting between them? (b) How many excess electrons are on each drop, giving it its charge imbalance?

Homework Equations



My question relates to part B, the excess electrons

## q=ne ##

e=1.602 x 10-19 C (elementary charge value from textbook)

In this case e is negative because it is the charge on an electron.

The Attempt at a Solution



## n = \frac{q}{ -e}##

##n = \frac{-1.00\times 10^{-16} C}{-1.602\times 10^{-19} C}##

n = 624.2 electrons ∴ there are fractional number of electrons but we can only have integers for our n values.

Plugging in e = 1.60 x 10-19, we get 625 electrons. This is the answer listed in the book.

My questions is when do we use the 1.602 x 10-19 value and when do we use the 1.60 x 10-19? Is this just something the author was not paying attention to when he wrote the book and made up the numbers?

I know it may be a trivial question but this is driving me nuts. Thanks for helping out.
 
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ChiralWaltz said:

Homework Statement



Two tiny, spherical water drops, with identical charges of -1.00 x 1016 C, have a center-to-center separation of 1.00 cm.(a) What is the magnitude of the electrostatic force acting between them? (b) How many excess electrons are on each drop, giving it its charge imbalance?

Homework Equations



My question relates to part B, the excess electrons

## q=ne ##

e=1.602 x 10-19 C (elementary charge value from textbook)

In this case e is negative because it is the charge on an electron.

The Attempt at a Solution



## n = \frac{q}{ -e}##

##n = \frac{-1.00\times 10^{-16} C}{-1.602\times 10^{-19} C}##

n = 624.2 electrons ∴ there are fractional number of electrons but we can only have integers for our n values.

Plugging in e = 1.60 x 10-19, we get 625 electrons. This is the answer listed in the book.

My questions is when do we use the 1.602 x 10-19 value and when do we use the 1.60 x 10-19? Is this just something the author was not paying attention to when he wrote the book and made up the numbers?

I know it may be a trivial question but this is driving me nuts. Thanks for helping out.

The question provides data accurate to 2 decimal places, i.e 'identical charges of -1.00 x 1016 C'.

It is therefore reasonable to use the 1.60x10-19 figure since you can't really determine an answer to a greater accuracy than what was given to you.

This is a general rule of thumb
 
Thanks you BOAS.

So any time there is a constant in the calculation, we use the amount of significant figures from the given data to determine how many sig figs we can use from the constant.

I'm so happy it was as simple issue as sig figs.
 

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