A) \[\text{12}\text{.5 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-2}}}\text{J}\]
B) \[\text{-25 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-2}}}\text{J}\]
C) \[\text{-12}\text{.5 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-2}}}\text{J}\]
D) \[\text{25 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-2}}}\text{J}\]
Correct Answer: D
Solution :
From the formula charge on the capacitor is given by \[=CV=50\times {{10}^{-6}}\times 100=5\times {{10}^{-3}}C\] When the separation between the plates is doubled the capacity becomes half \[i.e.,\left( \frac{e}{Z} \right)\]and potential becomes 2V. Now change in energy \[=\frac{1}{2}\left( \frac{C}{2} \right){{(2V)}^{2}}-\frac{1}{2}C{{V}^{2}}=\frac{1}{2}C{{V}^{2}}\] So, energy spent\[=\frac{1}{2}\times 50\times {{10}^{-6}}\times {{(100)}^{2}}\] \[=25\times {{10}^{-2}}\,Joule\]
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