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Step-by-Step Detailed Solution
Step 1: Determine the initial charge on the charged capacitor
The capacitor of capacitance C = 60\,\text{pF} is charged to a voltage V = 20\,\text{V} . The charge stored on this charged capacitor is
Q = CV = (60 \times 10^{-12}\,\text{F}) \times (20\,\text{V}) = 1.2 \times 10^{-9}\,\text{C}.
Step 2: Find the initial energy stored
The energy stored initially in the charged capacitor is given by
U_{\text{initial}} = \tfrac{1}{2} C V^{2}.
Substituting the values, we get
U_{\text{initial}} = \tfrac{1}{2}\times 60 \times 10^{-12}\,\text{F} \times (20\,\text{V})^{2}.
U_{\text{initial}} = \tfrac{1}{2} \times 60 \times 10^{-12} \times 400 \,\text{J}
= 1.2 \times 10^{-8}\,\text{J}
= 12\,\text{nJ}.
Step 3: Connect the charged capacitor to the uncharged capacitor
The charged 60\,\text{pF} capacitor is disconnected from the supply and then connected in parallel to an identical (but initially uncharged) 60\,\text{pF} capacitor. When connected in parallel, the total capacitance becomes
C_{\text{total}} = 60\,\text{pF} + 60\,\text{pF} = 120\,\text{pF}.
Step 4: Determine the common voltage across both capacitors
Since no external supply is connected after they are joined, charge is conserved. The total charge in the system remains Q = 1.2 \times 10^{-9}\,\text{C}. The common voltage V_{f} across each capacitor after sharing is found by
V_{f} = \dfrac{Q}{C_{\text{total}}} = \dfrac{1.2 \times 10^{-9}\,\text{C}}{120 \times 10^{-12}\,\text{F}} = 10\,\text{V}.
Step 5: Calculate the final energy stored and the energy lost
The final energy stored in both capacitors combined is
U_{\text{final}} = \tfrac{1}{2} C_{\text{total}} (V_{f})^2.
Substitute the values:
U_{\text{final}} = \tfrac{1}{2} \times 120 \times 10^{-12}\,\text{F} \times (10\,\text{V})^{2}
= \tfrac{1}{2} \times 120 \times 10^{-12} \times 100 \,\text{J}
= 6 \times 10^{-9}\,\text{J}
= 6\,\text{nJ}.
Thus, the energy lost during the redistribution of charge is
U_{\text{lost}} = U_{\text{initial}} - U_{\text{final}} = 12\,\text{nJ} - 6\,\text{nJ} = 6\,\text{nJ}.
