How many electrons have been removed from a positively charged electroscope?

AI Thread Summary
To determine how many electrons have been removed from a positively charged electroscope with a net charge of 7.5x10^-11 C, one must divide the net charge by the charge of a single electron, which is 1.6x10^-19 C. This calculation leads to the conclusion that the number of electrons removed is approximately 4.69x10^8. The participants confirm that this method is correct and express understanding of the calculation process. The discussion effectively clarifies the relationship between charge and the number of electrons.
mimictt
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Hello everyone! I'm having a problem with this question from my book...

How many electrons have been removed from a positively charged electroscope if it has a net charge o 7.5x10^-11 C?

Can someone explain to me how to solve this problem>?
Thank you very much!
 
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Hello,

What is the charge of a single electron? What about two? What about an arbitrary number call it "x" electrons?
 


What's the charge on each electron?
 


it's 1.6x10^-19 C

so does it mean that 7.5x10^-11/ 1.6x10^-19 is the number of electrons?
 


oh i think i got it! thank you so much!
 


mimictt said:
it's 1.6x10^-19 C

so does it mean that 7.5x10^-11/ 1.6x10^-19 is the number of electrons?

yes!
 
Thread 'Variable mass system : water sprayed into a moving container'

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}...
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