A parallel plate capacitor is charged from a cell and then isolated from it. The separation between the plates is now increased

Options

A. The force of attraction between the plates will decrease.
B. The field in the region between the plates will not change.
C. The energy stored in the capacitors will increase.
D. The potential difference between the plates will decrease.

Answer

B, C

Explanation

b.,c.
`C=(Q)/(V)` Also, `C=(epsilon_(0)A)/(d)`
`As` `d` increases, `C` will decrease. Therefore, `V` should increase. As `Q` remains the same.
`E=(V)/(d)=(Q)/(Aepsilon_(0))` will be constant.
Energy, `U=(Q^(2))/(2C)`. As `C` decreaes, `U` will increase.