ilikebigboottys
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How would I do part b?

http://prntscr.com/ncxu9b
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ilikebigboottys
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bumppp
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Callicious
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(Original post by ilikebigboottys)
bumppp
Bumppp >: )

Okidoke.

I assume you've done the first part fine.

For the second part, the charge on the capacitor remains constant as you move the plates apart.

Thus, charge is "conserved" as it were.

Separation is constant.

As you remove the dielectric from the capacitor, the relative permittivity of the space within the capacitor reduces, and as such, the capacitance decreases.

If the capacitance decreases, but charge remains constant, what quantity increases, can you form an equation to solve for that?

(Sorry for the dirty reply, but, it should help! I'm off to bed, it's 00:38 and I literally have a mountain load to study. There is literally a meter high mountain of books, on my bed. That's how much there is to do.)

Another Edit
I assume that you know the equation relating permittivity and the capacitance for a plane-plate conducting capacitor like this one...

If not, then it's safe to assume that the electric field strength within the capacitor and hence the potential difference across it is inversely proportional to the relative permittivity K, which you have been given.
Last edited by Callicious; 1 year ago
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