AP EAMCET · PHYSICS · Capacitance
One of the two identical capacitors having the same capacitance C, is charged to a potential \(\mathrm{V}_1\) and the other is charged to a potential \(\mathrm{V}_2\). If they are connected with their like plates together, then the decrease in the electrostatic potential energy of the combined system is
- A \(\frac{\mathrm{C}}{4}\left(\mathrm{~V}_1^2-\mathrm{V}_2^2\right)\)
- B \(\frac{C}{4}\left(V_1^2+V_2^2\right)\)
- C \(\frac{C}{4}\left(V_1-V_2\right)^2\)
- D \(\frac{\mathrm{C}}{4}\left(\mathrm{~V}_1+\mathrm{V}_2\right)^2\)
Answer & Solution
Correct Answer
(C) \(\frac{C}{4}\left(V_1-V_2\right)^2\)
Step-by-step Solution
Detailed explanation
\(U_{initial} = \frac{1}{2}CV_1^2 + \frac{1}{2}CV_2^2\) \(Q_{total} = CV_1 + CV_2\) \(V_{final} = \frac{CV_1 + CV_2}{C+C} = \frac{V_1 + V_2}{2}\) \(U_{final} = \frac{1}{2}(2C)\left(\frac{V_1 + V_2}{2}\right)^2 = \frac{C}{4}(V_1 + V_2)^2\)…
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