| A parallel plate capacitor is charged to a potential difference of 100 V and disconnected from the source of emf. A slab of dielectric is then inserted between the plates. Which of the following three quantities change? |
| (i) The potential difference |
| (ii) The capacitance |
| (iii) The charge on the plates |
A) only (i) and (ii)
B) only (i) and (iii)
C) only (ii) and (iii)
D) All (i), (ii) and (iii)
Correct Answer: A
Solution :
The charge of the plates cannot change because the plates are isolated. Inserting the dielectric changes the capacitance and therefore changes the potential difference (from \[V=\frac{Q}{C}\]), since the charge is fixed.
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