question_answer

The work done in placing the dielectric slab inside one of the capacitors as shown in diagram

A) \[\frac{C{{V}^{2}}}{2}\left( \frac{K-1}{K+1} \right)\]

B)        \[\frac{C{{V}^{2}}}{4}\left( \frac{K-1}{K+1} \right)\]

C) \[\frac{C{{V}^{2}}}{4}\left( \frac{K+1}{K-1} \right)\]

D)        \[\frac{C{{V}^{2}}}{2}\left( \frac{K+1}{K-1} \right)\]

Correct Answer: B

Solution :

[b]   \[{{U}_{i}}=\frac{1}{2}(C/2){{V}^{2}}=\frac{C{{V}^{2}}}{4}\] \[{{U}_{f}}=\frac{1}{2}\left( \frac{C\times KC}{C+KC} \right){{V}^{2}}=\frac{KC{{V}^{2}}}{2(1+K)}\] \[W=\Delta U={{U}_{f}}-{{U}_{i}}\] \[=\frac{KC{{V}^{2}}}{2(1+K)}-\frac{C{{V}^{2}}}{4}=\frac{C{{V}^{2}}}{2}\left[ \frac{K}{1+K}-\frac{1}{2} \right]\] \[=\frac{C{{V}^{2}}}{2}\left[ \frac{2K-1-K}{2(1+K)} \right]=\frac{C{{V}^{2}}}{4}\left( \frac{K-1}{K+1} \right)\]