A parallel plate capacitor has a charge Q of 5.5 × 10^7 C on one plate and 5.5 ×

Question

A parallel plate capacitor has a charge Q of 5.5 × 10^7 C on one plate and 5.5 × 10^7 C on the other. (a) The distance between the plates is increased by 50% while the charge on each plate stays the same. What happens to the capacitance and the energy stored in the capacitor? (b) Make a “before and after” sketch indicating the change in Q, d, and V. (c) The distance between the plates is increased by 50% while the capacitor is connected to a battery. What happens to the capacitance and the energy stored in the capacitor? (d) Make a “before and after” sketch indicating the change in Q, d, and V.

Explanation / Answer

capacitance is inversely proportional to the distance between plates

so if the distance is increased capacitance will decrease

a ) here Q remains same but distance is increased by 50 %

if initially we had C it changed to 0.5C ie. Capacitance is halfed

initially energy stored was 0.5CV^2 or 0.5Q^2/C

it will be changed to 0.5Q^2/0.5C = Q^2/C

so energy stored is doubled

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b) initial

Q , V , C ,d

final conditons are Q , 2V , 0.5 C , 2d __________________________________________________________________________________

c) even here the distance is increased by 50% but being connected to a battery

so Q is not constant but V is .

by increasing distance C is reduced to 0.5C

energy stored will be 0.5CV^2 initially it will be changed to 0.5(0.5C)V^2

so energy is halfed and so is Capacitance

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d) initial conditions

C , V , d , Q

final conditions are

0.5C , V , 2d , 0.5Q

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