A capacitor is charged with a battery and then it's plate separation is increased without disconnecting the battery .what will be the change in a) energy stored in the capacitor b) potential difference across the plates of the capacitor d) electric field between the plates of the capacitor e) energy density stored in the capacitor.
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Answers
Well there will be two cases .
We are going to use the formulas below.
1) C= A × epsilon/ d
2) F= QE
3) V= E.d
4) Q=CV where :
Q= charge on plates
F= force between plates
V= voltage across plates
E= electric field between the plates
C= capacitance of the plates
d= distance between the plates.
CASE 1: Capacitor is charged and supply voltage is disconnected.
As d increases electric fied ( E) remains same but voltage(V) goes up from above formula. But capacitance (C) decreases so Q remains constant and hence from 2nd formula Force will be constant.
CASE 2 : Capacitor is charged and it is still connected to supply in that case V is constant(it has to be constant).
So as d increases E and C decreases from above formula, as a result Q too decreases ( Q= CV) and hence force decreases.
While increasing the separation of the plate without disconnecting the battery,
1. Stored energy in the capacitor will decrease by E = 1/2cv^2.
2. Potential difference would remain the same which is equal to emf of the cell.
3. As the distance increase and the potential difference remains the same, electric field would decrease.
4. Energy density would decrease.