Electrostatics Question 136
Question: A parallel plate capacitor of capacity $ C _{0} $ is charged to a potential $ V _{0} $ (i) The energy stored in the capacitor when the battery is disconnected and the separation is doubled $ E _{1} $ (ii) The energy stored in the capacitor when the charging battery is kept connected and the separation between the capacitor plates is doubled is $ E _{2}. $ Then $ E _{1}/E _{2} $ value is [EAMCET 2003]
Options:
A) 4
B) 3/2
C) 2
D) 1/2
Show Answer
Answer:
Correct Answer: A
Solution:
Let $ E=\frac{1}{2}C _{0}V _{0}^{2}\text{then}{E _{\text{1}}}=2E $
and $ E _{2}=\frac{E}{2} $
So $ \frac{E _{1}}{E _{2}}=\frac{4}{1} $
