A) \[\frac{1}{2}C{{V}^{2}}\left( \frac{k-1}{k+1} \right)\]
B) \[\frac{1}{2}C{{V}^{2}}\left( \frac{1-k}{k+1} \right)\]
C) \[\frac{1}{2}C{{V}^{2}}(k-1)\]
D) \[\frac{1}{2}C{{V}^{2}}(1-k)\]
Correct Answer: B
Solution :
As the battery is disconnected, the charge on the plates remains the same. The capacitance after the dielectric is inserted is \[C'=\frac{(A/2){{\varepsilon }_{0}}}{d}+\frac{(A/2){{\varepsilon }_{0}}k}{d}\] \[=\frac{A{{\varepsilon }_{0}}}{d}\left( \frac{k+1}{2} \right)=C\left( \frac{k+1}{2} \right)\] The work done is \[W={{U}_{f}}-{{U}_{i}}=\frac{{{Q}^{2}}}{2C'}-\frac{{{Q}^{2}}}{2C}=\frac{{{Q}^{2}}}{2C}\left[ \frac{2}{k+1}-1 \right]\] \[=\frac{{{(CV)}^{2}}}{2C}\left( \frac{1-k}{k+1} \right)=\frac{1}{2}C{{V}^{2}}\left( \frac{1-k}{k+1} \right)\] Hence, the correction option is [b].
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