A 60μF capacitor is charged to store 0.3J of energy. An uncharged 120μF capacitor is then connected in parallel with the first one through a perfectly conducting lead. What is the final energy of the system?

Perfect conducting lead

Energy stored in 60μF capacitor is charged and contains 0.3J of energy. To obtain voltage in the circuit, I need to transpose fallowing formulae:

\[

W=\frac{1}{2}\times{}C\times{}V^2

\]

\[

0.3=\frac{1}{2}\times{}60\times{}{10}^{-6}\times{}V^2

\]

\[

\frac{0.3\times{}2}{60\times{}{10}^{-6}}=V^2

\]

\[

\sqrt{1000}=V

\]

\[

V=100V

\]

Assume that, the capacitor 60μF is fully charged, the voltage in the circuit will be equal to 100V.

To calculate final energy in the system after connection in parallel 120μF capacitor, I have to add both capacitance of capacitors and apply to formula:

\[

W=\frac{1}{2}CV^2

\]

\[

W=\frac{{10}^{-6}\times{}(60+120)\times{}{100}^2}{2}

\]

\[

W=0.9J

\]

Answer:

The total energy in the system is equal to 0.9J.

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