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} $



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