Question: Capacitor miscellany

here are my questions about capacitor:

1 #51 b)Why doubling?

The energy stored in a capacitor is given by:

U=(1/2)QV=(1/2)CV²=(1/2)Q²/C.

In Ch. 17, #51 (b) we are told that the amount of charge on each plate is doubled (ie. Q is doubled), and the capacitor remains connected to a battery (ie. V is held constant). So, we can look at an initial energy, U₀, and a new energy after the charge is doubled, U':

U₀=Q₀V
U'=Q'V=(2Q₀)V

The ratio of the new energy to the old is then:

U'/U₀=Q'V/Q₀V=2

So, indeed if the charge on the capacitor is doubled, the energy stored in the capacitor is doubled.

2 Say, we have 2+Q on the left plate and 2 -Q on the rite, can I just add 1 +Q to the left and add nothing to the rite? Or it is a must for me to add the same amount of charges to both sides?

You must add the same amount of charge to both sides. (Check out the Ch. 17 summary on where it describes a capacitor.) If you added more of one charge to one side than the other, the net charge of the capacitor would be non-zero. I'm not sure what exactly would happen, but it wouldn't be a simple capacitor anymore and wouldn't follow the behaviour we've been discussing. Sorry I can't give a better explanation than that.

3 If I added 1 +Q to left and 1 -Q to rite, the total charge should be +2Q, rite?

Well, the net charge would be zero, the total amount of charge would be |2Q| (those || are to show we are talking magnitude/absolute value and disregarding the signs), but the charge on the capacitor (the "Q" you use in the capacitor equation) would be 1Q. The measure of charge that matters in a capacitor is how much you have separated off to each plate, so it is this charge you consider.

4 #77b) what i did was PE/2=1/2*m*v^2, what is wrong with my solution( i know it must be wrong -_-||)?

I had a student ask this question last year, I think the answer given there will help you also. Check out electrostatics: previous student questions, it is the second item on the page.

I believe your problem lies in the factor of 1/2 you are using for the potential energy... I'm not sure where this comes from, and I'm pretty sure it isn't necessary. ;)

Good Luck, hope this helps. As always you can ask follow-up questions.