A capacitor C is connected to a battery of V volts and is fully charged.Keeping
ID: 1326031 • Letter: A
Question
A capacitor C is connected to a battery of V volts and is fully charged.Keeping the battery connected, the spacing between the capacitor plates is reduced to half.
What happens to the potential difference between the two plates? Why?
What happens to the charge on the capacitor? Why?
What happens to the energy stored in the capacitor? Why?
Now the battery is disconnected, and then the plate spacing is restored to its original value.
What happens to the potential difference between the two plates? Why?
What happens to the charge on the capacitor? Why?
What happens to the energy stored in the capacitor? Why?
Note: It is not enough if you say
Explanation / Answer
1) potential difference between two plates remain same since it is in parallel to the battery
2) q= cv and c = e0 A/d d becomes half so c becomes double so q alsobecomed double
3) energy stored U = eo A V^2 / d so on reducing d to half energy doubled
4) potential difference as V = Qd/ ??A so now doubling the d will double the potential difference
5) charge still remains same since it is indepedent on spacing as Q= CV and C becomes half while V becomes double so net no change
6) As the charges on the two plates are opposite they attract which means that external work needs to be done to increase their separation. This work increases the potential energy of the charges
U = eo A V^2 / d
and V is doubled and d also so net is U doubled since V2 beomes 4 times
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