A) \[\frac{125\times {{10}^{-3}}}{2}\]
B) \[\frac{12.5\times {{10}^{-3}}}{2}\]
C) \[\frac{1.25\times {{10}^{-3}}}{2}\]
D) \[\frac{0.125\times {{10}^{-3}}}{2}\]
Correct Answer: A
Solution :
Initially energy stored in capacitor \[{{E}_{1}}=\frac{1}{2}C{{V}^{2}}\] \[=\frac{1}{2}\,\times 100\times {{10}^{-6}}\times 50\times 50\] \[\Rightarrow \] \[{{E}_{1}}=125\times {{10}^{-3}}J\] When the distance is doubled, the capacitance decreases to 1/2, i.e., \[C'=50\,\mu F\] \[\therefore \] Final energy, \[{{E}_{2}}=\frac{1}{2}C'{{V}^{2}}=\frac{1}{2}\times 50\times {{10}^{-6}}\times {{(50)}^{2}}\] \[=\frac{125\times {{10}^{-3}}}{2}\] Additional energy, \[E={{E}_{1}}-{{E}_{2}}=\frac{125\times {{10}^{-3}}}{2}J\]You need to login to perform this action.
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