3 concentric conducting spheres, the outer one connected to ground

AI Thread Summary
The discussion centers on the implications of connecting the outermost conducting sphere to ground, specifically regarding the potential and surface charge density. It is established that grounding implies V(r=R_5)=0, leading to considerations about the charge densities on the other surfaces. Participants discuss the relationships between charge densities on different surfaces and the electric field outside the outer sphere. The hints provided help clarify the relationships necessary to solve the problem effectively. Overall, the conversation aids in understanding the electrostatic behavior of the system.
Sokolov
Messages
14
Reaction score
1
Homework Statement
Three concentric conductor spheres, the outer of them connected to ground, have electrical charges[itex]Q_A[/itex], [itex]Q_B[/itex] and [itex]Q_C[/itex], and radii[itex]R_1[/itex], [itex]R_2[/itex], [itex]R_3[/itex], [itex]R_4[/itex] and[itex]R_5[/itex], from the inner to the outer. Find the induced charges and the equations of the electric field and the potential of the system.
Relevant Equations
[tex] \oint \vec E\cdot d\vec S =\frac{q}{\epsilon_0}[/tex]
Esferas.png

What would the fact that the fifth surface is connected to the ground imply: that V(r=R_5)=0 or that \sigma _5=0?
 
Physics news on Phys.org
Hello Sokolov, :welcome: !

Ground is considered ##V=0##.
Perhaps not extremely interesting for part two, since ##E## is a derivative.
 
  • Like
Likes Sokolov
Sokolov said:
Homework Statement:: Three concentric conductor spheres, the outer of them connected to ground, have electrical chargesQ_A, Q_B and Q_C, and radiiR_1, R_2, R_3, R_4 andR_5, from the inner to the outer. Find the induced charges and the equations of the electric field and the potential of the system.
Relevant Equations:: \oint \vec E\cdot d\vec S =\frac{q}{\epsilon_0}

View attachment 258000
What would the fact that the fifth surface is connected to the ground imply: that V(r=R_5)=0 or that \sigma _5=0?
Couple of hints to add to BvU's:
1. what can you say about ##\sigma_2 ## vs. ##\sigma_1 ##? Etc?
2. What must be the field outside R5? What does that imply for ## \sigma_5 ## assuming you've worked out the other sigmas?

P.S. ## \sigma_i ## means charge density on surface i.
 
  • Like
Likes Sokolov
BvU said:
Hello Sokolov, :welcome: !

Ground is considered ##V=0##.
Perhaps not extremely interesting for part two, since ##E## is a derivative.

Hi BvU, thanks for the answer! It was just what I needed to know in order to be able to solve the problem.
 
rude man said:
Couple of hints to add to BvU's:
1. what can you say about ##\sigma_2 ## vs. ##\sigma_1 ##? Etc?
2. What must be the field outside R5? What does that imply for ## \sigma_5 ## assuming you've worked out the other sigmas?

P.S. ## \sigma_i ## means charge density on surface i.

Thanks rude man! With your hints and BvU's answer I think that I have been able to solve the problem correctly :) .
 
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...
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}...
Back
Top