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1. Electric Charges and Fields
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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 $(iii)$
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 $\mathrm{V}=\mathrm{Q} / \mathrm{C}$ ), since the charge is fixed.
Standard 12
Physics
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