After charging a capacitor the battery is removed. Now by placing a dielectric slab between the plates :-
After charging a capacitor the battery is removed. Now by placing a dielectric slab between the plates :-
- A
The potential difference between the plates and the energy stored will decrease but the charge on plates will remain same
- B
the charge on the plates will decrease and the potential difference between the plates will increase
- C
the potential difference between the plates will increase and energy stored will decrease but the charge on the plates will remain same
- D
the potential difference, energy stored and the charge will remain unchanged.
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- [AIPMT 1999]
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Match the pairs
Capacitor
Capacitance
$(A)$ Cylindrical capacitor
$(i)$ ${4\pi { \in _0}R}$
$(B)$ Spherical capacitor
$(ii)$ $\frac{{KA{ \in _0}}}{d}$
$(C)$ Parallel plate capacitor having dielectric between its plates
$(iii)$ $\frac{{2\pi{ \in _0}\ell }}{{ln\left( {{r_2}/{r_1}} \right)}}$
$(D)$ Isolated spherical conductor
$(iv)$ $\frac{{4\pi { \in _0}{r_1}{r_2}}}{{{r_2} - {r_1}}}$
Match the pairs
| Capacitor | Capacitance |
| $(A)$ Cylindrical capacitor | $(i)$ ${4\pi { \in _0}R}$ |
| $(B)$ Spherical capacitor | $(ii)$ $\frac{{KA{ \in _0}}}{d}$ |
| $(C)$ Parallel plate capacitor having dielectric between its plates | $(iii)$ $\frac{{2\pi{ \in _0}\ell }}{{ln\left( {{r_2}/{r_1}} \right)}}$ |
| $(D)$ Isolated spherical conductor | $(iv)$ $\frac{{4\pi { \in _0}{r_1}{r_2}}}{{{r_2} - {r_1}}}$ |