Is Zero Electric Field on a Closed Surface Indicative of No Net Charge?

[willdefender]

QUESTIONS
5.The electric field E is zero at all points on a closed surface; is there necessarily no net charge within the surface? If a surface encloses zero net charge, is the electric field necessarily zero at all points on the surface?

6.Define gravitational flux in analogy to electric flux. Are there "sources" and "sinks" for the gravitational field as there are for the electric field?

8.A spherical basketball(a nonconductor) is given a charge Q distributed uniformly over its surface.What can you say about the electric field inside the ball? A person now steps on the ball, collapsing it, and forcing most of the air out without altering the charge. What can you say about the field inside now?

12.A conductor carries a net positive charge Q. There is a hollow cavity within the conductor, at whose center is a negative point charge -q. What is the charge on (a)the outer surface of the conductor and (b) the inner surface of the conductor?


ATTEMPT
5.----I think the first one is yes because according to Gauss's Law, if a surface enclose net charge, E at points on surface would not be 0; as for the last one, I think it is wrong because even though there is no sinks or sources in a closed surface, electric field lines are still allowed to pass through the surface, which results that E at some points on the surface is not 0.

6.----I think anything that has mass can be considered as "sinks", but how about "sources"?

8.----At first, the field is determined the Gauss's Law. After collapsing, the field doesn't change because of no loss of charge.

12.----I think both are 0. For the outer surface, the net charge it enclose is 0, so that E on the surface is 0.

 

[diazona]

5. Sounds good
6. Right; so what do you think about sources?
8. Yes, the field is determined by Gauss's law at first. So what can you say about its value inside the basketball? Afterwards, note that the basketball is no longer spherical.
12. Why would you say the net charge enclosed in the outer surface is 0? Think about this: what is the electric field in any region filled by conducting material?

 

[willdefender]

5. Sounds good
6. Right; so what do you think about sources?
8. Yes, the field is determined by Gauss's law at first. So what can you say about its value inside the basketball? Afterwards, note that the basketball is no longer spherical.
12. Why would you say the net charge enclosed in the outer surface is 0? Think about this: what is the electric field in any region filled by conducting material?

Thanks for your replying!
6.If something that has a negative mass, it can be considered as a source.

8.So if the charge are in the ball(not distributed on the ball), the charge will not changed.But does not the electric field determined only by the charge? Does the distribution of charge affect that?

12.I think the net charge is 0, so...
Is the electric field in any region filled by conducting material 0?

 

[diazona]

8. Yes, the electric field is determined by the charge distribution...

12. It's true that the electric field in any region filled by conducting material is 0. (Do you know why?) Now think about applying Gauss's law to a surface which runs through the conductor.

 

[willdefender]

8. Yes, the electric field is determined by the charge distribution...

12. It's true that the electric field in any region filled by conducting material is 0. (Do you know why?) Now think about applying Gauss's law to a surface which runs through the conductor.

8.We can calculate the previous field with E1=Q/(4*Π*ε0*r*r), after collasping, we should calculate it with E2=δ/(2*ε0）=E1/2, isn't it?

12. I think the charge on the outer surface is Q-q, while the one on the inner surface is q-Q.

 

[diazona]

8.We can calculate the previous field with E1=Q/(4*Π*ε0*r*r), after collasping, we should calculate it with E2=δ/(2*ε0）=E1/2, isn't it?

Inside the ball? That's not correct for E1. Or well, technically you could calculate it that way, but it would involve an integral that you don't really have to do. What happened to using Gauss's Law?

12. I think the charge on the outer surface is Q-q, while the one on the inner surface is q-Q.

Nope, I don't believe that's correct.