E&M Electrostatics 3 (urgent)

2011-05-02 5:42 am

1. A positively-metal sphere A of radius a is joined by a conducting wire to an unchanged metal sphere B of radius b placed far away from the first sphere. The ratio of the surface charge density on sphere A to that on sphere B is? (ans:b/a)
The solution to this question states that the electric potential of A and B are the same, but why?

2. An insulated metal sphere S of radius 2r carries a charge +Q. Another insulated metal sphere of radius r carrying a charge +2Q at infinity to S through a conducting wire. What would the charge on S be after removing the wire? (ans: +3Q/2)
Same as the above question, why are there electric potential the same when the two metal sphere are connected? And compared to the above question, are there any differences between the two set-up? I am not refering to their sizes, charges, I mean any differences that may lead to differene methods involed in soloving the question.

3. Please refer to this website:
http://www.lsforum.net/board/thread-170585-1-3.html (ans:C)
Can some1 explain option 1 and 3 to me? The answer key metions 'shielding effect' when explaining option 1, what is it and how is it related to it?

4. An additional question about wave: http://www.lsforum.net/board/thread-170990-1-2.html Why is the answer C? A displaces to the right so why it starts in negative?

thanks so much

更新1:

For 3, why 'But this is physically impossible on a conducting sphere.' ? Can you explain more please? thanks

回答 (1)

天同

2011-05-02 8:15 pm

✔ 最佳答案

1. The two spheres are at the same potential because they have been joined by a conductin wire. Should their potentials not be equal, charges would flow from one sphere to the other until the potential difference between the spheres is zero. (remeber the important concept that potential difference causes charge to flow, just like a "height difference" in mechanics causes a mass to fall).

2. The two situations are the same.The potential of the 1st sphere (of radius 2r) is at a potential lower than that of the 2nd sphere (of radius r). As such, when they are connected by a wire, cahrges flow from the 2nd sphere (because it is at a higher potential) to the first sphere because of the potential difference between them (remember the important concept mentioned in Q1).

Hence, the potential of the 1st sphere rises because it gains in charge, whereas the potential of the 2nd sphere falls because of loss in charge. The potential difference between the two spheres therefore gets narrowing down. The flow of charge stops when the potential of both spheres are equal. There is now no potential difference between the two spheres and thus no further charge flow.

3. I think only statement 1 is correct.
Statement 1 is apparent. If charge distribution is not even, the part of sphere with higher surface charge density would be at higher potential. There is then a potential difference, and hence electric field lines occur on the sphere surface. But this is physically impossible on a conducting sphere.
Statement 3 is similar to statemtn 1. In a conductor, there must not be any eelctric field line exist, and all points in a conductor are at the same potential.
4. Can't see the attachment. Sorry...can't help.

2011-05-02 18:46:26 補充：
Re: your supplementary question.
A conducting surface should be an equipotential surface. This indicates that the potential gradient is zero, and hence elecric field doesn't exist.

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