How to find the number of excess electrons?

  • Thread starter Thread starter pebbles
  • Start date Start date
  • Tags Tags
    Electrons
AI Thread Summary
To find the number of excess electrons on a ball with a charge of -4.00*10^-17 C, the charge per electron is 1.60*10^-19 C. The correct calculation involves dividing the total charge by the charge of one electron, resulting in 4.00*10^-17 C / 1.60*10^-19 C, which simplifies to 2.5*10^2 electrons. Understanding division with powers of ten is crucial, as 10^-17 divided by 10^-19 equals 100. The result confirms that excess charge must be a multiple of the elementary charge, supporting the calculation.
pebbles
Messages
95
Reaction score
0

Homework Statement



How many excess electrons are on a ball with a charge of -4.00*10^-17 C?

Homework Equations


I know that the charge per electron is 1.60 *10^-19C.


The Attempt at a Solution



My textbook does not explain how to do this, but I thought I would divide-->4.00*10^-17 C * 1 electron/-1.60*10^-19. I got -2.5*10^-36. The answer from the book is 2.5*10^2 electrons.
I did some messing around and did this-->4.00*10^-17C*1 e/1.60*10^19C=2.5*10^2 electrons.
What's the correct way to do this problem?
Thanks in advance.
 
Physics news on Phys.org
Your set-up is right and the units will check. But how do you divide

4.0 x 10^-17 / 1.6 x 10^-19 ?

What is 1 / 1.6 x 10^-19 ?

(In fact, your check is also incorrect. You may want to review how division works with powers of ten and what negative exponents mean. 10^-17 / 10^-19 = 100 ; 10^-17 / 10^19 = 10^-36 .)
 
uhh, I'm confused....?
 
You're dividing by 1.6 x .0000000000000000001 . So 4 / 1.6 is 2.5 , but what is

10^-17 / 10^-19 =

0.00000000000000001 / 0.0000000000000000001 ?
 
oh i see!
10^-17/10^-19=100
so therefore 2.5*10^2.
thanks so much! :D
 
Charge is quantized so the excess charge has to be a multiple of e (elementary charge)

q=ne, where n is the number of electrons.

That's why it works I believe since your textbook didn't explain it.
 
  • Like
Likes HRubss
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...

Similar threads

Back
Top