Electrostatics Question 76
Question: The plates of a parallel plate capacitor of capacity $ 50\mu C $ 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 [MP PET 1997; JIPMER 2000]
Options:
A) $ 25\times {{10}^{-2}}J $
B) $ -12.5\times {{10}^{-2}}J $
C) $ -25\times {{10}^{-2}}J $
D) $ 12.5\times {{10}^{-2}}J $
Show Answer
Answer:
Correct Answer: A
Solution:
$ W _{ext}=\frac{1}{2}C’V{{’}^{2}}-\frac{1}{2}CV^{2} $
$ =( \frac{1}{2} )( \frac{C}{2} ){{(2V)}^{2}}-\frac{1}{2}CV^{2}=\frac{1}{2}CV^{2} $
$ W _{ext}=\frac{1}{2}\times 50\times {{10}^{-6}}\times {{(100)}^{2}}=25\times {{10}^{-2}}J $
