Calculating Electrostatic Force and Charge Imbalance in Tiny Water Drops

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The electrostatic force between two identical charged water drops, each with a charge of -1.04 x 10^-16 C and separated by 1.5 cm, is calculated to be 4.32159 x 10^-19 N. To determine the number of excess electrons on each drop, the total charge is divided by the charge of a single electron, resulting in 650 excess electrons per drop. The calculations confirm the principles of electrostatics and charge quantization. Understanding these concepts is crucial for applications involving small charged particles. The discussion effectively clarifies the calculation methods for both the electrostatic force and charge imbalance.
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Two tiny, spherical water drops, with identical charges of -1.04 10-16 C, have a center-to-center separation of 1.5 cm.
(a) What is the magnitude of the electrostatic force acting between them?
(b) How many excess electrons are on each drop, giving it its charge imbalance?

Ok so, i know that the answer for A is 4.32159e-19.
However, I'm not sure how to do B.
I think its something like the total charge divided by a single charge. I'm not sure though.
 
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for b you are correct. All you do is divide the charge on the drop by the charge on an electron.
 
650.

Thanks you.
 
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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