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Chapter 24: Problem 20

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.

Short Answer

Expert verified
Answer: The energy stored in the capacitor increases after the insertion of the dielectric, and the dielectric is pulled into the space between the plates.

Step by step solution

01

Part a: Energy before dielectric insertion

The energy stored in a capacitor before the dielectric was inserted can be calculated using the formula \[U = \frac{1}{2}CV^2,\] where \(U\) is the stored energy, \(C\) is the capacitance, and \(V\) is the constant potential difference. Using this formula, we can calculate the initial energy stored as \[U_{initial} = \frac{1}{2}CV^2.\]
02

Part b: Energy after dielectric insertion

When the dielectric is inserted between the plates, the capacitance changes. The new capacitance, \(C'\), is given by \[C' = \kappa C,\] where \(\kappa\) is the dielectric constant. The energy stored after dielectric insertion can be calculated using the same formula as before: \[U_{final} = \frac{1}{2}C'V^2,\] By substituting \(C' = \kappa C\), we get \[U_{final} = \frac{1}{2}(\kappa C)V^2.\]
03

Part c: Dielectric interaction with the plates

To determine if the dielectric was pulled or pushed into the space between the plates, we should analyze the forces acting on it. The force can be determined by the change in energy when the dielectric is inserted. \[F = U_{final} - U_{initial}\] Substitute the expressions for \(U_{final}\) and \(U_{initial}\): \[F = \frac{1}{2}(\kappa C)V^2 - \frac{1}{2}CV^2\] Factor out \(\frac{1}{2}CV^2\): \[F = \frac{1}{2}CV^2(\kappa-1)\] Since \(\kappa > 1\) for dielectric materials, the value of \(F\) will be positive. This means that the energy stored in the capacitor has increased after inserting the dielectric. Therefore, the dielectric was pulled into the space between the plates.

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Most popular questions from this chapter

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?

Which of the following capacitors has the largest charge? a) a parallel plate capacitor with an area of \(10 \mathrm{~cm}^{2}\) and a plate separation of \(2 \mathrm{~mm}\) connected to a \(10-\mathrm{V}\) battery b) a parallel plate capacitor with an area of \(5 \mathrm{~cm}^{2}\) and a plate separation of \(1 \mathrm{~mm}\) connected to a \(10-\mathrm{V}\) battery c) a parallel plate capacitor with an area of \(10 \mathrm{~cm}^{2}\) and a plate separation of \(4 \mathrm{~mm}\) connected to a \(5-\mathrm{V}\) battery d) a parallel plate capacitor with an area of \(20 \mathrm{~cm}^{2}\) and a plate separation of \(2 \mathrm{~mm}\) connected to a \(20-\mathrm{V}\) battery e) All of the capacitors have the same charge.

A dielectric slab with thickness \(d\) and dielectric constant \(\kappa=2.31\) is inserted in a parallel place capacitor that has been charged by a \(110 .-\mathrm{V}\) battery and having area \(A=\) \(100, \mathrm{~cm}^{2},\) and separation distance \(d=2.50 \mathrm{~cm}\). a) Find the capacitance, \(C,\) the potential difference, \(V,\) the electric field, \(E,\) the total charge stored on the capacitor \(Q\), and electric potential energy stored in the capacitor, \(U\), before the dielectric material is inserted. b) Find \(C, V, E, Q,\) and \(U\) when the dielectric slab has been inserted and the battery is still connected. c) Find \(C, V, E, Q\) and \(U\) when the dielectric slab is in place and the battery is disconnected.

When working on a piece of equipment, electricians and electronics technicians sometimes attach a grounding wire to the equipment even after turning the device off and unplugging it. Why would they do this?

The potential difference across two capacitors in series is \(120 . \mathrm{V}\). The capacitances are \(C_{1}=1.00 \cdot 10^{3} \mu \mathrm{F}\) and \(C_{2}=1.50 \cdot 10^{3} \mu \mathrm{F}\) a) What is the total capacitance of this pair of capacitors? b) What is the charge on each capacitor? c) What is the potential difference across each capacitor? d) What is the total energy stored by the capacitors?
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