Electrostatics Question 648
Question: Two identical capacitors, have the same capacitance$ C $ . One of them is charged to potential $ V _{1} $ and the other to $ V _{2} $ the negative ends of the capacitors are connected together. When the positive ends are also connected, the decrease in energy of the combined system is
Options:
A) $ \frac{1}{4}C(V^{2} _{{{1}^{{}}}}-V^{2} _{2}) $
B) $ \frac{1}{4}C(V^{2} _{1}+V^{2} _{2}) $
C) $ \frac{1}{4}C{{(V _{1}-V _{2})}^{2}} $
D) $ \frac{1}{4}C{{(V _{1}+V _{2})}^{2}} $
Show Answer
Answer:
Correct Answer: C
Solution:
[c] Initial energy of the system $ U _{i}=\frac{1}{2}CV^{2} _{1}+\frac{1}{2}CV^{2} _{2} $
When the capacitors are joined, common potential $ V=\frac{CV _{1}+CV _{2}}{2C}=\frac{V _{1}+V _{2}}{2} $
Final energy of the system $ U _{f}=\frac{1}{2}(2C)V^{2}=\frac{1}{2}2C{{( \frac{V _{1}+V _{2}}{2} )}^{2}}=\frac{1}{4}C{{(V _{1}+V _{2})}^{2}} $
Decrease in energy = $ U _{i}-U _{f}=\frac{1}{4}C{{(V _{1}-V _{2})}^{2}} $
