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A parallel capicitor is made of two plates with cross sectional area A which are

ID: 2239005 • Letter: A

Question

A parallel capicitor is made of two plates with cross sectional area A which are separated by a distance d (with a vacuum between the plates). The capacitor is connected to a battery until a charge Q is stored on it and a total potential difference VC is measured across the plates (the potential energy stored at this time is UC). Situation I: The capacitor is disconnected from the battery, and its plates moved so the distance between them is doubled. a. What is the potential difference measured across the plates now (as a multiple of the original potential difference)? VC b. What is the potential energy stored in the capacitor now (as a multiple of the original potential energy stored)? UC Situation II: Instead of the above, the capacitor is left connected to the battery, and its plates moved so the distance between them is divided by 2. c. What is the potential difference measured across the plates now (as a multiple of the original potential difference)? VC d. What is the potential energy stored in the capacitor now (as a multiple of the original potential energy stored)? UC

Explanation / Answer

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C = eA / d

Vc = Q / C

Uc = 0.5 * C * Vc^2

Situation 1 : In this case the charge will remain constant but the Potential will vary.

a) Charge will remain constant, Q' = Q

Capacitance, C' = eA/d'

here d' = d/4

So, C' = 4C

So, V' = Q'/C' = Q/4C = Vc /4


b) U' = 0.5 * C' * V'^2 = 0.5 * (4C) * (Vc/4)^2 = Uc/4


Situation 2 :

Here the voltage source is same, so voltage will remain same, but more charge will get deposited

c) V' = Vc since it is connected to same source of battery

d) C' = C/4

So, U' = 0.5 * C' * V'^2 = 0.5 * (C/4) * Vc^2 = = Uc / 4


Answers:

VC/4

UC/4

VC

UC/4

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