Counting electrons: current and charge

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SUMMARY

The discussion centers on calculating the charge and number of electrons passing through a flashlight bulb with a steady current of 0.50 A over 2.0 minutes. The correct calculation yields a charge of 60 C and approximately 3.8 x 1020 electrons. The confusion arises from the use of incorrect values in calculations, particularly the charge of an electron, which is 1.6 x 10-19 C. Proper understanding of current, charge, and electron calculations is essential for accurate results.

PREREQUISITES
  • Understanding of electric current (I) and its units (Amperes)
  • Knowledge of charge (q) and its calculation from current and time
  • Familiarity with the charge of an electron (1.6 x 10-19 C)
  • Basic proficiency in using scientific calculators for exponential calculations
NEXT STEPS
  • Review the principles of electric current and charge calculations
  • Learn about the relationship between current, charge, and time using the formula q = It
  • Explore the concept of elementary charge and its significance in physics
  • Practice using scientific calculators for exponential and scientific notation calculations
USEFUL FOR

Students studying physics, particularly those focusing on electricity and magnetism, as well as educators looking to clarify concepts related to current and charge calculations.

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Hi everyone. I thought I understood this problem, but now I'm unsure. Everything is worked out step by step with the answers, but when I try to duplicate it, I get something different. Can anyone shed some light on it?

Homework Statement



Suppose there is a steady current of 0.50 A in a flashlight bulb lasting for 2.0 min. How much charge passes through the bulb during this time? How many electrons does this represent?

The current and time elapsed are given; therefore, the definition of current allows the calculation of the charge q. Since each electron carries a charge of magnitude 1.6x10-19 C, q can be converted into the number of electrons.


Homework Equations



Given:

I = 0.50 A
t = 2.0 min = 1.2x102 s

I=q/t, so the magnitude of the charge is

q = It = (0.50A)(1.2x102 s) = (0.50C/s)(1.2x102 s) = 60 C

Solving for number of electrons from q = ne

n = q/e = 60 C/1.6x10-19 C/electrons = 3.8x1020 electrons


The Attempt at a Solution



I understand the first part of the problem (everything to get the 60 C). I've done several of these problems before and this is the first one that's tripped me up. I'm using a calculator to do this and when I enter that into my computer, I come up with 3.75x10-18.

I know that 1.6x10-19 C is the electric charge of an electron (although, earlier in the book it states that it's -1.6x10-19 C and that protons have an electric charge of 1.6.x10-19 C). The only way I can get the answer they came up with is if I use 1.6x1019 and that doesn't make any sense to me. What am I missing?
 
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Your answers look right to me, 60C and 3.8x10^20 electrons. You my not be operating the calculator right? Do simple math problems that you know the answer to to figure out the calculator or go online and get the manual for your calculator?

Hope this helps.
 

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