After charging a capacitor the battery is removed. Now by placing a dielectric slab between the plates :-
The potential difference between the plates and the energy stored will decrease but the charge on plates will remain same
the charge on the plates will decrease and the potential difference between the plates will increase
the potential difference between the plates will increase and energy stored will decrease but the charge on the plates will remain same
the potential difference, energy stored and the charge will remain unchanged.
A parallel plate capacitor of capacitance $5\,\mu F$ and plate separation $6\, cm$ is connected to a $1\, V$ battery and charged. A dielectric of dielectric constant $4$ and thickness $4\, cm$ is introduced between the plates of the capacitor. The additional charge that flows into the capacitor from the battery is........$\mu C$
Two identical parallel plate capacitors are connected in series to a battery of $100\,V$. A dielectric slab of dielectric constant $4.0$ is inserted between the plates of second capacitor. The potential difference across the capacitors will now be respectively
Write the formula of capacitance of capacitor having dielectric constant $\mathrm{K} = 2$.
Between the plates of a parallel plate capacitor a dielectric plate is introduced just to fill the space between the plates. The capacitor is charged and later disconnected from the battery. The dielectric plate is slowly drawn out of the capacitor parallel to the plates. The plot of the potential difference across the plates and the length of the dielectric plate drawn out is
A capacitor is half filled with a dielectric $(K=2)$ as shown in figure A. If the same capacitor is to be filled with same dielectric as shown, what would be the thickness of dielectric so that capacitor still has same capacity?