Physics problem

Superman

2009-04-27 12:21 am

P and Q are two conducting spheres each carrying positive charges. P is suspended from a non-conducting thread. Q is supported by a non-conducting stand. They rest in equilibrium. A positively charged rod is brought close to Q from the right-hand side. P rises a little bit and the net charge on P remains unchanged.

Now Q is earthed for a very short while, which of the following statements is/are true?

A. P remains stationary

B. P falls and sticks to Q

C. P falls, collides with Q and bounces off from Q. Finally, P rests in a lower position than before.

D. P falls, collides with Q and bounces off from Q. This repeats with the amplitude of swing of P decresasing until P sticks to Q.

Explain your answer.

更新1:

electric field lines tend to distribute evenly and symmetrically. Besides they tend to have the shortest path. 唔係好明白咩意思..

更新2:

補充呢個係第2條問題, 唔關上面事...

回答 (1)

天同

2009-04-27 5:37 am

✔ 最佳答案

The answer is option A
I suppose P and the charged rod are on the opposite side of Q.
When the charged rod is placed near to Q, it induces more positive charge on the side of Q near to P. Since like charges repell, P will be pushed away a little bit by the induced charges.
Since the charge on P is given to remain unchanged when Q is earthed, the same amount of positive charge on Q is being attracted by P and hence earthing has no effect on these charges. Hence, P will remain stationary.
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I think you refer to the electric field line pattern from a point charge of spherical charge.
These are in fact physical properties of field lines. Due to isotropy, electric field lines from a +ve point charge is directed radially outward. The spacings between field lines at the same distance from the charge are the same.
Field lines actually behave like "stretched elastic bands", which tend to shorten as far as they could. That's why they always tened to have the shortest path on which they would have the shortest length.

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