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Chapter 17: Problem 110

A parallel plate capacitor is attached to a battery that supplies a constant voltage. While the battery is still attached, a dielectric of dielectric constant \(\kappa=3.0\) is inserted so that it just fits between the plates. What is the energy stored in the capacitor after the dielectric is inserted in terms of the energy \(U_{0}\) before the dielectric was inserted?

Short Answer

Expert verified
Answer: The energy stored in the parallel plate capacitor after the dielectric is inserted is 3.0 times the initial stored energy, or 3.0U₀.

Step by step solution

01

Calculate the initial energy stored in the capacitor

To find the initial energy stored in the capacitor, we will use the formula \(U_{0} = \frac{1}{2} CV^{2}\), where \(C\) is the capacitance and \(V\) is the voltage. The problem does not give us specific values for \(C\) and \(V\), so we will leave the formula as it is for now.
02

Determine the new capacitance after the dielectric is inserted

After the dielectric is inserted, the capacitance of the parallel plate capacitor will change. We can find the new capacitance, \(C_{new}\), using the formula \(C_{new} = \kappa C\), where \(\kappa\) is the dielectric constant. The dielectric constant is given as \(\kappa = 3.0\), so the new capacitance is \(C_{new} = 3.0C\).
03

Calculate the energy stored in the capacitor after the insertion of the dielectric

Now, we will find the new energy stored in the capacitor, \(U_{new}\), using the same formula we used in Step 1, but with the new capacitance, \(C_{new}\). So, the formula for the new energy stored in the capacitor is \(U_{new} = \frac{1}{2} C_{new}V^{2}\). Substitute the expression for \(C_{new}\) from Step 2: \(U_{new} = \frac{1}{2}(3.0C)V^{2}\).
04

Express the new energy stored in terms of the initial energy

We want to express the new energy, \(U_{new}\), in terms of the initial energy, \(U_{0}\). From Step 1, we know that \(U_{0} = \frac{1}{2} CV^{2}\), so \(CV^{2} = 2U_{0}\). Substituting this expression into our formula for \(U_{new}\) from Step 3, we get: \(U_{new} = \frac{1}{2}(3.0C)V^{2} = \frac{1}{2}(3.0)(2U_{0}) = (3.0)U_{0}\). The energy stored in the parallel plate capacitor after the dielectric is inserted is \(3.0U_{0}\).

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Most popular questions from this chapter

Suppose a uniform electric field of magnitude \(100.0 \mathrm{N} / \mathrm{C}\) exists in a region of space. How far apart are a pair of equipotential surfaces whose potentials differ by \(1.0 \mathrm{V} ?\)
Two parallel plates are \(4.0 \mathrm{cm}\) apart. The bottom plate is charged positively and the top plate is charged negatively, producing a uniform electric field of \(5.0 \times 10^{4} \mathrm{N} / \mathrm{C}\) in the region between the plates. What is the time required for an electron, which starts at rest at the upper plate, to reach the lower plate? (Assume a vacuum exists between the plates.)
Find the electric potential energy for the following array of charges: charge \(q_{1}=+4.0 \mu \mathrm{C}\) is located at \((x, y)=(0.0,0.0) \mathrm{m} ;\) charge \(q_{2}=+3.0 \mu \mathrm{C}\) is located at \((4.0,3.0) \mathrm{m} ;\) and charge \(q_{3}=-1.0 \mu \mathrm{C}\) is located at (0.0,3.0) \(\mathrm{m}\)
A parallel plate capacitor is charged by connecting it to a \(12-V\) battery. The battery is then disconnected from the capacitor. The plates are then pulled apart so the spacing between the plates is increased. What is the effect (a) on the electric field between the plates? (b) on the potential difference between the plates?
A capacitor can be made from two sheets of aluminum foil separated by a sheet of waxed paper. If the sheets of aluminum are \(0.30 \mathrm{m}\) by $0.40 \mathrm{m}$ and the waxed paper, of slightly larger dimensions, is of thickness \(0.030 \mathrm{mm}\) and dielectric constant \(\kappa=2.5,\) what is the capacitance of this capacitor?
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