Electrostatics Question 44
Question: A $ 6\mu F $ capacitor is charged from $ 10\ volts $ to$ 20\ volts $ . Increase in energy will be [CPMT 1987, 97; BCECE 2004]
Options:
A) $ 18\times {{10}^{-4}}J $
B) $ 9\times {{10}^{-4}}J $
C) $ 4.5\times {{10}^{-4}}J $
D) $ 9\times {{10}^{-6}}J $
Show Answer
Answer:
Correct Answer: B
Solution:
$ \Delta E=E _{Final}-E _{Initial}=\frac{1}{2}C(V _{Final}^{2}-V _{Initial}^{2}) $
$ =\frac{1}{2}\times 6\times (20^{2}-10^{2})\times {{10}^{-6}} $
$ =3\times (400-100)\times {{10}^{-6}}=3\times 300\times {{10}^{-6}}=9\times {{10}^{-4}}J $
