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