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

Answer: (D)

$\text{25 }\\!\\!\times\\!\\!\text{ 1}{{\text{0}}^{\text{-2}}}\text{J}$

Answer 2

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 $=\frac{1}{2}\times 50\times {{10}^{-6}}\times {{(100)}^{2}}$ $=25\times {{10}^{-2}}\,Joule$