Electrostatics Question 109

Question: A body of capacity $ 4\mu F $ is charged to $ 80V $ and another body of capacity $ 6\mu F $ is charged to 30V. When they are connected the energy lost by $ 4\mu F $ capacitor is [EAMCET 2001]

Options:

A) 7.8 mJ

B) 4.6 mJ

C) 3.2 mJ

D) 2.5 mJ

Show Answer

Answer:

Correct Answer: A

Solution:

Initial energy of body of capacitance 4 mF is $ U _{i}=\frac{1}{2}\times (4\times {{10}^{-6}}){{(80)}^{2}}=0.0128J $

Final potential on this body after connection is $ V=\frac{4\times 80+6\times 30}{4+6}=50V. $

So final energy on it $ U _{f}=\frac{1}{2}\times 4\times {{10}^{-6}}{{(50)}^{2}}=0.005J $

Energy lost by this body $ =U _{i}U _{f}=\text{ 7}.\text{8}mJ $



NCERT Chapter Video Solution

Dual Pane