91Ó°ÊÓ

To solve this problem, we need to determine the potential energy of the charged capacitor in both situations, which can be calculated using the formula for the energy stored in a capacitor: \(U = \frac{1}{2}CV^2\) where \(U\) is the potential energy, \(C\) is the capacitance, and \(V\) is the voltage across the capacitor.

When the capacitor remains connected to the charging battery, the voltage across the capacitor remains constant as the plates are separated. In this case, the potential energy can be calculated using the formula: \(U_{connected} = \frac{1}{2}CV_{battery}^2\)

When the capacitor is disconnected from the charging battery, the charge on the capacitor plates remains constant as the plates are separated. In this case, we can use the definition of capacitance as \(C = \frac{Q}{V}\), where \(Q\) is the charge on the plates. Rearranging this expression, we have: \(V_{disconnected} = \frac{Q}{C}\) Since the charge remains constant, the potential energy in this case can be calculated by substituting the constant charge formula for the voltage in the energy formula: \(U_{disconnected} = \frac{1}{2}C\left(\frac{Q}{C}\right)^2\)

Now that we have expressions for the potential energy in both situations, we can compare them to determine which requires more work. We can compare \(U_{connected}\) and \(U_{disconnected}\) by taking their ratio: \(\frac{U_{connected}}{U_{disconnected}} = \frac{\frac{1}{2}CV_{battery}^2}{\frac{1}{2}C\left(\frac{Q}{C}\right)^2} = \frac{V_{battery}^2}{\left(\frac{Q}{C}\right)^2}\) Since \(V_{battery} = \frac{Q}{C}\), the ratio simplifies to: \(\frac{U_{connected}}{U_{disconnected}} = 1\) Because the ratio is equal to 1, it takes the same amount of work to separate the plates of a charged parallel plate capacitor, whether it remains connected to the charging battery or after it has been disconnected from the charging battery.