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Chapter 17: Problem 112

A 4.00 - \(\mu \mathrm{F}\) air gap capacitor is connected to a \(100.0-\mathrm{V}\) battery until the capacitor is fully charged. The battery is removed and then a dielectric of dielectric constant 6.0 is inserted between the plates without allowing any charge to leak off the plates. (a) Find the energy stored in the capacitor before and after the dielectric is inserted. [Hint: First find the new capacitance and potential difference.] (b) Does an external agent have to do positive work to insert the dielectric or to remove the dielectric? Explain.

Short Answer

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Explain why.

Step by step solution

01

Calculate the initial energy stored in the capacitor (U₀)

First, we'll find the energy stored in the capacitor when charged by the battery, before inserting the dielectric. The formula for the energy stored in a capacitor is given by: \( U = \frac{1}{2}CV^2 \) where \(U\) is the energy stored, \(C\) is the capacitance, and \(V\) is the potential difference across the capacitor. Given that the initial capacitance is \(4.00\ \mu\mathrm{F}\) and the potential difference is \(100.0\ \mathrm{V}\), we can find the initial energy stored (U₀): \( U₀ = \frac{1}{2}(4.00\ \mu\mathrm{F})(100.0\ \mathrm{V})^2 \)
02

Calculate the new capacitance (C')

After the dielectric is inserted, the capacitance of the capacitor changes. The relationship between the initial capacitance (C₀) and the new capacitance (C') is given by the equation: \( C' = KC₀ \) where \(K\) is the dielectric constant. Given the dielectric constant is \(6.0\), the new capacitance can be calculated as: \( C' = 6.0 \times 4.00\ \mu\mathrm{F} \)
03

Calculate the new potential difference (V')

The potential difference across the capacitor changes when the dielectric is inserted, as the charge on the plates remains constant. The relationship between the initial voltage (V₀), the new voltage (V'), and the capacitance values is given by the equation: \( V₀ C₀ = V' C'\) Rearranging this equation, we can calculate the new potential difference: \( V' = \frac{V₀ C₀}{C'}\)
04

Calculate the energy stored in the capacitor with the dielectric (U')

Now that we have the new capacitance (C') and the new potential difference (V'), we can find the energy stored in the capacitor after the dielectric is inserted (U') using the energy formula: \( U' = \frac{1}{2}C' V'^2\)
05

Determine if positive work is required to insert or remove the dielectric

In order to determine if positive work is required to insert or remove the dielectric, we can compare the initial energy stored in the capacitor (U₀) with the final energy stored in the capacitor (U'). If the initial energy is greater than the final energy, an external agent has to do positive work to insert the dielectric. If the final energy is greater than the initial energy, an external agent has to do positive work to remove the dielectric. Otherwise, no external work is necessary.

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Most popular questions from this chapter

A parallel plate capacitor has a capacitance of \(1.20 \mathrm{nF}\) There is a charge of \(0.80 \mu \mathrm{C}\) on each plate. How much work must be done by an external agent to double the plate separation while keeping the charge constant?
A spherical conductor of radius \(R\) carries a total charge Q. (a) Show that the magnitude of the electric field just outside the sphere is \(E=\sigma / \epsilon_{0},\) where \(\sigma\) is the charge per unit area on the conductor's surface. (b) Construct an argument to show why the electric field at a point \(P\) just outside any conductor in electrostatic equilibrium has magnitude \(E=\sigma / \epsilon_{0},\) where \(\sigma\) is the local surface charge density. [Hint: Consider a tiny area of an arbitrary conductor and compare it to an area of the same size on a spherical conductor with the same charge density. Think about the number of field lines starting or ending on the two areas.]
A point charge \(q=+3.0\) nC moves through a potential difference $\Delta V=V_{\mathrm{f}}-V_{\mathrm{i}}=+25 \mathrm{V} .$ What is the change in the electric potential energy?
The bottom of a thundercloud is at a potential of $-1.00 \times 10^{8} \mathrm{V}\( with respect to Earth's surface. If a charge of \)-25.0 \mathrm{C}$ is transferred to the Earth during a lightning strike, find the electric potential energy released. (Assume that the system acts like a capacitor-as charge flows, the potential difference decreases to zero.)
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