Value of Charge Q: Why Use Distance not Radius?

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The discussion centers on a physics problem involving two charged volleyballs and the calculation of the electrostatic force between them. The confusion arises from the use of "radius" versus "distance" in Coulomb's law. It is clarified that in this context, "radius" refers to the distance between the two point charges, not half the distance or the length of the tethering strings. The correct approach involves using the full distance between the charges to accurately calculate the electrostatic force. Understanding this distinction resolves the initial misunderstanding about the terms used in the problem.
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Homework Statement



this is problem 1.4 from purcells electricity and magnetism 2nd edition.

Two volleyballs, mass 0.3 kg each, tethered by nylon strings and charged by an electrostatic generator, hang as shown in the diagram. which cam be found here http://www.fysik.su.se/grulab//LabDataBas/2005/2005_9_27_14_21_10/R1_03.pdf
the height is 2.5 m and distance between the charges 0.5 m the diagram is a little messed up.
What is the charge on each in coulombs assuming the charges are equal? When trying to work this problem, I ended up just looking up the solution online here. i was getting close to the answer. I used the radius instead pf the distance between the two charges. Hence my question:

I don't understand why the distance between the two charges is used. Should the radius not be used?
 
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random_user13 said:
I don't understand why the distance between the two charges is used. Should the radius not be used?
Used for what? For finding the electrostatic force between them? And by radius you mean the length of of the string?
 
Radius is the same thing as distance. In Coulomb's force equation, radius refers to the distance between the two point charges, not half the distance (like the definition of radius you are thinking).
 
That makes sense now. Definitely mixed up how that was being used.
 
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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