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Step-by-Step Solution
Step 1: Identify the Initial Energy
Suppose we have a capacitor of capacitance $C$ charged to a voltage $V$ by a battery. The energy stored in this charged capacitor is given by the formula
$$U_i = \frac{1}{2} C V^2 \quad \text{(Initial energy)}$$
Step 2: Connect the Uncharged Capacitor in Parallel
After the battery is disconnected, another identical capacitor (also of capacitance $C$) that is uncharged is connected in parallel with the charged capacitor:
Since the two capacitors are now connected in parallel and have the same capacitance, the total capacitance becomes $2C$. However, because charge is conserved (and the second capacitor was initially uncharged), the final common voltage will be shared equally among the two identical capacitors.
Step 3: Determine the Common Voltage
Let the initial charge on the first capacitor be $Q = CV$. When connected in parallel with an uncharged capacitor (also of capacitance $C$), the total charge $Q$ remains the same, but is now distributed between the two capacitors (which must be at the same final voltage $V'$). Hence:
$$V' = \frac{\text{Total charge}}{\text{Total capacitance}} = \frac{CV}{C + C} = \frac{CV}{2C} = \frac{V}{2}.$$
Step 4: Calculate the Final Energy
After connection in parallel, the combined system of the two capacitors behaves like a single equivalent capacitor of $2C$ charged to $V' = \frac{V}{2}$. Therefore, the final stored energy $U_f$ is
$$U_f = \frac{1}{2} \left(2C\right) \left(\frac{V}{2}\right)^2 = \frac{1}{2} \times 2C \times \frac{V^2}{4} = \frac{1}{4} C V^2.$$
Step 5: Compare the Initial and Final Energies
Originally, the energy was $U_i = \frac{1}{2} C V^2$. After connection in parallel, the final energy is $U_f = \frac{1}{4} C V^2$. Hence,
$$U_f = \frac{1}{2} U_i,$$
which shows that the energy is halved. Therefore, the total electrostatic energy of the system decreases by a factor of 2.
Conclusion
When an initially charged capacitor is connected in parallel with an identical uncharged capacitor, the total energy stored in the combined system becomes half of the initial energy. Hence, the total electrostatic energy of the resulting system
decreases by a factor of 2.
