Calculating potential on surface of sphere

AI Thread Summary
A spherical drop of water with a charge of 30 pC has a potential of 500 V at its surface. The potential is calculated using the equation Potential = (1/4εpi) q/r, leading to the conclusion that the charge can be treated as concentrated on the surface for the purpose of calculating potential. The discussion raises questions about whether the charge distribution affects the electric field outside the sphere, referencing Gauss's Law. Ultimately, it is established that for points outside the sphere, the charge can be considered as if it were located at the center, regardless of its actual distribution. This understanding is crucial for solving problems involving electric potential and fields around charged spheres.
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Homework Statement



A spherical drop of water carrying a charge of 30 pC has a potential of 500 V at its surface (which V = 0 at infinity). What is the radius of the drop?

Homework Equations



Potential = (1/4εpi) q/r


The Attempt at a Solution



When I first looked at this problem, I assumed that pure water would have charge evenly distributed, since it's not a conductor...and I was confused, because I wasn't sure how to calculate the charge on the surface, which is where the Voltage is given. To answer this problem, do we have to assume that the charge is located on the surface and then use the equation for potential? I did so, and got the right answer. However, I'm not sure whether I am right that we have to make this assumption, or whether there is some larger theoretical issue whereby it doesn't matter whether the charge is on the surface of evenly distributed throughout the sphere.
 
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mm2424 said:

Homework Statement



A spherical drop of water carrying a charge of 30 pC has a potential of 500 V at its surface (which V = 0 at infinity). What is the radius of the drop?

Homework Equations



Potential = (1/4εpi) q/r


The Attempt at a Solution



When I first looked at this problem, I assumed that pure water would have charge evenly distributed, since it's not a conductor...and I was confused, because I wasn't sure how to calculate the charge on the surface, which is where the Voltage is given. To answer this problem, do we have to assume that the charge is located on the surface and then use the equation for potential? I did so, and got the right answer. However, I'm not sure whether I am right that we have to make this assumption, or whether there is some larger theoretical issue whereby it doesn't matter whether the charge is on the surface of evenly distributed throughout the sphere.
What does Gauss's Law tell you about the E-field external to the drop. Does it matter whether the charge is distributed uniformly on the surface or distributed uniformly throughout the volume?
 
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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