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Chapter 17: Problem 76

A large parallel plate capacitor has plate separation of \(1.00 \mathrm{cm}\) and plate area of \(314 \mathrm{cm}^{2} .\) The capacitor is connected across a voltage of \(20.0 \mathrm{V}\) and has air between the plates. How much work is done on the capacitor as the plate separation is increased to $2.00 \mathrm{cm} ?$

Short Answer

Expert verified
Question: Calculate the work done on a parallel plate capacitor when its plate separation is increased from 1.00 cm to 2.00 cm while keeping the voltage constant at 20 V. The area of the plates is 314 cm² and the vacuum permittivity is 8.85 × 10⁻¹² F/m. Answer: The work done on the capacitor is -2.774 × 10⁻⁹ J.

Step by step solution

01

Find the initial capacitance

The formula for capacitance of a parallel plate capacitor is: \(C = \frac{\epsilon_0 A}{d}\) Where \(C\) is the capacitance, \(\epsilon_0\) is the vacuum permittivity, \(A\) is the area of the plates, and \(d\) is the distance between the plates. Using the given values, we can find the initial capacitance. \(\epsilon_0 = 8.85 × 10^{-12} \mathrm{F/m}\) \(A = 314 \mathrm{cm}^2 \times 10^{-4} \mathrm{m}^2/\mathrm{cm}^2 = 0.0314\,\mathrm{m}^2\) \(d_1 = 1.00\, \mathrm{cm} \times 10^{-2} \mathrm{m/cm} = 0.010\,\mathrm{m}\) \(C_1 = \frac{\epsilon_0 A}{d_1} = \frac{8.85 × 10^{-12} \mathrm{F/m} \times 0.0314\,\mathrm{m}^2}{0.010\,\mathrm{m}} = 2.774 \times 10^{-11}\,\mathrm{F}\)
02

Find the final capacitance

Now, we can find the final capacitance when the plate separation is increased to \(2.00\,\mathrm{cm}\). \(d_2 = 2.00\, \mathrm{cm} \times 10^{-2} \mathrm{m/cm} = 0.020\,\mathrm{m}\) \(C_2 = \frac{\epsilon_0 A}{d_2} = \frac{8.85 × 10^{-12} \mathrm{F/m} \times 0.0314\,\mathrm{m}^2}{0.020\,\mathrm{m}} = 1.387 \times 10^{-11}\,\mathrm{F}\)
03

Calculate the energy stored in the capacitor before and after

The energy stored in a capacitor can be calculated using the formula: \(U = \frac{1}{2}CV^2\) We can calculate the energy stored in the capacitor initially and after the plate separation is increased: \(U_1 = \frac{1}{2}C_1V^2 = \frac{1}{2}(2.774 \times 10^{-11}\,\mathrm{F})(20.0\,\mathrm{V})^2= 5.548 \times 10^{-9}\,\mathrm{J}\) \(U_2 = \frac{1}{2}C_2V^2 = \frac{1}{2}(1.387 \times 10^{-11}\,\mathrm{F})(20.0\,\mathrm{V})^2= 2.774 \times 10^{-9}\,\mathrm{J}\)
04

Calculate the work done on the capacitor

The work done on the capacitor can be found by taking the difference between the final and initial energies stored in the capacitor: \(W = U_2 - U_1 = 2.774 \times 10^{-9}\,\mathrm{J} - 5.548 \times 10^{-9}\,\mathrm{J} = -2.774 \times 10^{-9}\,\mathrm{J}\) The negative sign indicates that work is done by the capacitor, meaning that the energy is transferred from the capacitor to an external system. This is the work done on the capacitor as the plate separation is increased to \(2.00\,\mathrm{cm}\).

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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 positively charged oil drop is injected into a region of uniform electric field between two oppositely charged, horizontally oriented plates spaced $16 \mathrm{cm}$ apart. If the electric force on the drop is found to be $9.6 \times 10^{-16} \mathrm{N}\( and the potential difference between the plates is \)480 \mathrm{V}$ what is the magnitude of the charge on the drop in terms of the elementary charge \(e\) ? Ignore the small buoyant force on the drop.
A parallel plate capacitor is connected to a \(12-\mathrm{V}\) battery. While the battery remains connected, the plates are pushed together so the spacing is decreased. What is the effect on (a) the potential difference between the plates? (b) the electric field between the plates? (c) the magnitude of charge on the plates?
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 parallel plate capacitor is connected to a battery. The space between the plates is filled with air. The electric field strength between the plates is \(20.0 \mathrm{V} / \mathrm{m} .\) Then, with the battery still connected, a slab of dielectric \((\kappa=4.0)\) is inserted between the plates. The thickness of the dielectric is half the distance between the plates. Find the electric field inside the dielectric.
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