A capacitor has some dielectric between its plates, and the capacitor is connected to a DC source. The battery is now disconnected and then the dielectric is removed. State whether the capacitance, the energy stored in it, electric field, charge stored and the voltage will increase ro remain constant.

Explanation

The battery or DC source charges the capacitor. When the battery is disconnected, there will be no flow of charge and hence charge on the capacitor will remain constant even if the capacitance is changed. Let the capacitance of the given capacitor be C, the dielectric constant of the dielectric slab inside the capacitor be K, the voltage across the plates of the capacitor be V, the charge on the capacitor be Q and the electric field in the region between the plates be E.
Now, on removing the dielectric, the capacitance C will decrease to `C.=(C )/(K)`.
Energy stored by the capacitor is given by `U=(Q^(2))/(2C)`.
As we know that the charge stored by the capacitor will not change even if its capacitance changes. Thus, the energy stored by the capacitor, voltage across its plates will change according to capacitance.
So, after removing the dielectric the energy stored in it will become:
`U.=(Q^(2))/(2C.)=(Q^(2))/(2C//K)`
`rArr U.=KU`
Voltage across the plates of the capacitor is given by
`V=(Q)/(C )`
After removing the dielectric the voltage will be
`V.=(Q)/(C.)=(Q)/(C//K)=KV`
Electric field between the plates of the capacitor is given by `E.=(V.)/(d)=(KV)/(d)=KE`
Therefore, the capacitance will decrease, the energy stored in capacitor, the electric field and the voltage across the capacitor will increase, and charge stored by the capacitor will remain constant in the given situation.