Charge flow of grounded sphere?

AI Thread Summary
A conducting spherical shell with no net charge contains a +7 Coulombs charge inside. Initially, the shell has a -7 Coulombs charge on its inner surface and +7 Coulombs on its outer surface. When the shell is grounded, electrons flow from the ground to neutralize the positive charge, resulting in a net charge of -7 Coulombs on the shell after disconnection. This conclusion confirms the charge flow and equilibrium achieved through grounding. The final net charge on the spherical shell is indeed -7 Coulombs.
yeahhyeahyeah
Messages
29
Reaction score
0

Homework Statement


A conducting spherical shell , which has no net charge, has a charged object inside the shell somewhere with charge=+7 Coulombs.

Later the conducting spherical shell is connected to ground and whatever charge flows does so and a new equilibrium charge on the shell is reached. The shell is now again disconnected from ground. What now is the net total charge, if any, on the spherical shell?

The Attempt at a Solution



OK initially, before grounding, the shell would have a net charge of 0, a charge of +7 coulombs on its outer surface and a charge of -7 coulombs on its inner surface, right?

Once it's grounded... electrons will flow from the ground to the shell (because the outer surface is +7 so positive) , then, once it is removed, the shell will now have a net charge of -7 coulombs

Is this right?
 
Last edited:
Physics news on Phys.org
Perfectly correct.
 
thanks!
 
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}...
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top