1.

An air capacitor is connected to a battery for a long time so that the capacitor is fully charged. Now the battery is disconnected. Then the following actions take place one after another. X - Separation between the plates is increased to double Y - A dielectric slab is inserted to occupy entire space between the plates Z - The battery is connected again

Answer»

`{:(P,Q,R,S),(1,2,3,4):}`
`{:(P,Q,R,S),(4,3,1,2):}`
`{:(P,Q,R,S),(4,3,2,1):}`
`{:(P,Q,R,S),(2,4,1,3):}`

SOLUTION :In case of an isolated capacitor, the stored charge remains constant and hence the electric field intensity, which is given as `E=(Q)/(epsilon_(0)A)`, remains the same after the process X. Potential difference between the plates is given by V = Ed. Therefore, in the same electric field, when separation between the plates is doubled then the potential difference is ALSO doubled after X.
When the dielectric SLAB is inserted then we know that charge remains constant because the capacitor is still not connected with the battery. Due to insertion of the slab, the capacitance increases. Hence energy stored, which is given by `U=(Q^(2))/(2C)`, decreases due to an increasein the capacitance. Hence energy stored in the capacitor decreases after Y. But after the process Z, when the battery is reconnected, further charge is supplied by the battery due to an increased capacity. Hence, the energy stored in the capacitor increases and this energy comes from work done by the battery.


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