# 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. Plzz answr fast

The capacitance of a parallel plate capacitor is given by $C=\frac{\epsilon A}{d}$, where A is the area and d is the separation of the plates. Now since the capacitor is connected with the cell the potential will not change. But capacity will decrease if we increase the separation.

a) Energy will decrease as the energy is given by $E=\frac{1}{2}c{v}^{2}$
b) Potential will remain same and will be equal to the emf of cell.
c) Electric field will decrease since the potential difference will remain same and distance increasess.
d) Energy density will also decreases since the energy is decreases.

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