# How many electrons are there

#### Leikrod

Joined Oct 10, 2017
16
1 amp = 6,250,000,000,000,000,000 electrons per second
1 amp = 6.25 x 10^18 electrons per second
1 amp = 1 coulomb

Question 4

How many electrons are there in 2 ×10−5 coulombs of charge?
There are approximately 1.25 ×1014 electrons in this quantity of charge.

Hello , there is something wrong with my math because I cant get that answer, would anybody mind telling me the steps for it?

#### AlbertHall

Joined Jun 4, 2014
11,311
1 ampere second = 1 coulomb

#### Leikrod

Joined Oct 10, 2017
16
1 ampere second = 1 coulomb

I divided 1/6.25 x10^18 and multiplied the answer times .00002 to get 3.2 x10^-25 but thats not even close to the answer they r giving so I'm definitely doing something wrong

#### jpanhalt

Joined Jan 18, 2008
11,088

1 A does not equal 1 C.

#### Leikrod

Joined Oct 10, 2017
16

#### jpanhalt

Joined Jan 18, 2008
11,088
Several members here are of generations that learned to do calculations with the units. Your response in Post #3 doesn't show the units. You ask what is wrong? The answer, show the units and you will see.

1 C = 6.25x10^18 electrons

You are asked, "How many electrons are there in 2 ×10−5 coulombs of charge? " Example: If a kg of grapes has 100 grapes, how many grapes are in 0.1 kg?
Answer: 10 grapes (100 grapes/kg x 0.1 kg = 10 grapes as grapes/kg x kg = grapes/kg x kg = grapes ).

Why did you see the need to divide "1" by "6.25X18^18?"

#### dl324

Joined Mar 30, 2015
12,871
The actual steps to the answer would be appreciated
Do you know how to use the factor label method?

#### crutschow

Joined Mar 14, 2008
27,213
I divided 1/6.25 x10^18 and multiplied the answer times .00002 to get 3.2 x10^-25
How did you arrive at that calculation?

#### MrChips

Joined Oct 2, 2009
23,528
Learn to check your units in calculations.
This is called dimensional analysis.

#### WBahn

Joined Mar 31, 2012
26,398
I divided 1/6.25 x10^18 and multiplied the answer times .00002 to get 3.2 x10^-25 but thats not even close to the answer they r giving so I'm definitely doing something wrong
Well, let's do what you claimed to do and see what the math tells us.

$$\frac{1}{6.25 \times 10^{18} \frac{electrons}{coulomb}} \times 0.00002 coulombs \; = \; 3.2 \times 18^{-24} \frac{coulombs^2}{electron}$$

We all make silly mistakes when setting up equations or when working the math. You will make them your entire career. One hallmark of a professional is to utilize every reasonable means at their disposal to prevent making mistakes, render as many of the mistakes that do get made readily detectable, and to proactively correct errors once they are found.

Most of these silly mistakes will mess up the units -- whether the units are there to get messed up or not. But IF you religiously track your units throughout your work, they WILL be there to get messed up and you will see that they are messed up usually immediately, which will allow you to quickly spot and fix the mistake and move on.

The alternative is to ignore the units and just tack on the units you WANT the answer to have onto the end and assume that you didn't make any mistakes. That is how airliners have run out of fuel in midflight and how hundred million dollar space probes have been slammed into planets. Or, in this case, you have something to compare your answer to (which is seldom the case in the real world -- after all, engineers are hired to solve peoples' problems and if those people already had the answer, they wouldn't be hiring an engineer in the first place) you end up at a loss to explain it and must ask strangers on a forum to find your mistake.

I'm not putting you down -- at least you bothered to ask (many students don't and just go on without a second thought). But hopefully you can see that a very simple, easy, and free habit can save you a lot of trouble. Consider just how much time you would have saved on this one problem if you could have seen that your units didn't work out compared to the amount of time and effort required to start your thread and wait for others to respond.

People that refuse to track their units throughout their work often state that they don't have time to waste on doing so -- yet they always seem to have time to rework problems over and over or retake courses when they fail because they can't get the correct answers. I've seen it time and again. Tracking your units will SAVE you SO much time it is hard to overestimate its value. Then consider that in the real work an engineer making a mistake that doesn't get caught (but could have had they tracked their units) can have catastrophic consequences -- incompetent doctors can kill people, but are usually limited to doing so one at a time; incompetent engineers can kill people in job lots.

#### SLK001

Joined Nov 29, 2011
1,548
Why don't you just sit down and count them?

#### Leikrod

Joined Oct 10, 2017
16
Thank you all, I was able to solve it. 2 x10^-5 Coulombs x 6.25 x10^18 electrons/1 Coulombs = 1.25 x10^14 electrons.

#### Leikrod

Joined Oct 10, 2017
16
I'm not putting you down -- at least you bothered to ask (many students don't and just go on without a second thought). But hopefully you can see that a very simple, easy, and free habit can save you a lot of trouble. Consider just how much time you would have saved on this one problem if you could have seen that your units didn't work out compared to the amount of time and effort required to start your thread and wait for others to respond

No worries, your explanation was enough. I simply did not have the knowledge to correctly perform the exercise, but now I do.

#### dl324

Joined Mar 30, 2015
12,871
1 amp = 6.25 x 10^18 electrons per second
1 amp = 1 coulomb

After correcting the error in the second equation you get:
$$\small 2 * 10^{-5} C * \frac{6.25 * 10^{18} electrons}{1 C} = 1.25 * 10^{14} electrons$$

#### WBahn

Joined Mar 31, 2012
26,398
Note that, strictly speaking, the answer is actually -1.24 x 10^14 electrons in 20uC of charge (meaning that you need to remove that many electrons to get that charge).