Question

The plates of a parallel plate capacitor of capacity 50 $ \mu F $ are charged to a potential of 100 volts and then separated from each other so that the distance between them is doubled. How much is the energy spent in doing so?

• 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} $

Answer 1

The Correct Option is D

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 $