A parallel plate capacitor is charged and then the battery is disconnected. When the plates of the capacitor are brought closer, then

Options

A. energy stored in the capacitor decreases
B. the potential differences between the plates decreases
C. the capacitance increases
D. the electric field between the plates decreases

Answer

A, B, C

Explanation

If the battery is disconected, then `q=`constant
`C=(epsilon_0A)/d or Cprop1/d`
d is decreased. Hence, C will increase.
`U=1/2q^2/c or U prop 1/C`
C is increasing.Hence U will decrease.
`V=q/C or Vprop1/C`
`C` is increasing. Hence, `V` will decrease.