Calculating Net Charge on a Metal Sphere with Added Electrons

AI Thread Summary
The net charge on a metal sphere initially charged at +6 µC is calculated after adding 1 x 10^14 electrons. Each electron has a charge of -1.6 x 10^-19 C, resulting in a total negative charge of -1.6 x 10^-5 C from the electrons. The initial charge of +6 µC converts correctly to +6 x 10^-6 C. Adding the charges gives a net charge of 2.2 x 10^-5 C, which is incorrect due to a miscalculation in converting the initial charge. The correct conversion maintains the +6 µC as +6 x 10^-6 C without changing the digits.
osulongboard
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Homework Statement


A Metal sphere has a charge of +6uC. What is the net charge after 1*10^14 electrons have been placed on it?



Homework Equations


Single Electron - -1.6*10^-19


The Attempt at a Solution



1st. Convert +6uC to 1.6*10^-6 (is this right?)
2nd. (-1.6*10^-19)(1*10^14)=1.6*10^-5
3rd. (6*10^-6)+(1.6*10^-5)= 2.2*10^-5 (?)


I don't think this is a very hard problem but i want to make sure I am doing the basics right. Thanks in Advance!
 
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osulongboard said:
1st. Convert +6uC to 1.6*10^-6 (is this right?)

No. Changing "micro" to 10^-6 is like moving a decimal point, without changing the digits. You can't change the 6 to a 1.6
 
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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