| ASSERTION: As battery remains connected, potential of capacitor will not change. |
| REASON: If the capacitor is kept connected with the battery and then the dielectric slab is inserted between the plates, both the charge on capacitor and the energy stored in the capacitor gets decreased. |
A) If both assertion and reason are true and reason is the correct explanation of assertion.
B) If both assertion and reason are true and reason is not the correct explanation of assertion.
C) If assertion is true but reason is false.
D) If both assertion and reason are false.
E) If assertion is false but reason is true.
Correct Answer: C
Solution :
On inserting a dielectric slab the capacitance of capacitor \[{{C}_{0}}\] is increased to\[C\left( =\,\,K{{C}_{0}} \right)\], Therefore charge \[{{q}_{0}}\left( =\,{{C}_{0}}\,V \right)\]will increases to \[q\left( =\,K{{q}_{0}} \right)\] Initial potential energy, \[{{U}_{0}}=\frac{1}{2}{{C}_{0}}\,{{V}^{2}}\]. Hence, final potential energy, \[U=\frac{1}{2}C{{V}^{2}}=\frac{1}{2}K\,{{C}_{0}}\,{{V}^{2}}=K\,{{U}_{0}}\].
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