Does it take more work to separate the plates of a charged parallel plate capacitor while it remains connected to the charging battery or after it has been disconnected from the charging battery?

Short Answer

Expert verified
Answer: 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.

Step by step solution

01

Identifying the relevant formulas

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.
02

Determine the situation when the capacitor is connected to the battery

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: \(U_{connected} = \frac{1}{2}CV_{battery}^2\)
03

Determine the situation when the capacitor is disconnected from the battery

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: \(U_{disconnected} = \frac{1}{2}C\left(\frac{Q}{C}\right)^2\)
04

Comparing the potential energies

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.

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Most popular questions from this chapter

A parallel plate capacitor is constructed from two plates of different areas. If this capacitor is initially uncharged and then connected to a battery, how will the amount of charge on the big plate compare to the amount of charge on the small plate?

A parallel plate capacitor is connected to a battery for charging. After some time, while the battery is still connected to the capacitor, the distance between the capacitor plates is doubled. Which of the following is (are) true? a) The electric field between the plates is halved. b) The potential difference of the battery is halved. c) The capacitance doubles. d) The potential difference across the plates does not change. e) The charge on the plates does not change.

A parallel plate capacitor of capacitance \(C\) is connected to a power supply that maintains a constant potential difference, \(V\). A close-fitting slab of dielectric, with dielectric constant \(\kappa\), is then inserted and fills the previously empty space between the plates. a) What was the energy stored on the capacitor before the insertion of the dielectric? b) What was the energy stored after the insertion of the dielectric? c) Was the dielectric pulled into the space between the plates, or did it have to be pushed in? Explain.

A large parallel plate capacitor with plates that are square with side length \(1.00 \mathrm{~cm}\) and are separated by a distance of \(1.00 \mathrm{~mm}\) is dropped and damaged. Half of the areas of the two plates are pushed closer together to a distance of \(0.500 \mathrm{~mm}\). What is the capacitance of the damaged capacitor?

Two parallel plate capacitors, \(C_{1}\) and \(C_{2},\) are connected in series to a \(96.0-\mathrm{V}\) battery. Both capacitors have plates with an area of \(1.00 \mathrm{~cm}^{2}\) and a separation of \(0.100 \mathrm{~mm} ;\) \(C_{1}\) has air between its plates, and \(C_{2}\) has that space filled with porcelain (dielectric constant of 7.0 and dielectric strength of \(5.70 \mathrm{kV} / \mathrm{mm}\) ). a) After charging, what are the charges on each capacitor? b) What is the total energy stored in the two capacitors? c) What is the electric field between the plates of \(C_{2} ?\)

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