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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 cell membrane has a surface area of \(1.0 \times 10^{-7} \mathrm{m}^{2},\) a dielectric constant of \(5.2,\) and a thickness of \(7.5 \mathrm{nm}\) The membrane acts like the dielectric in a parallel plate capacitor; a layer of positive ions on the outer surface and a layer of negative ions on the inner surface act as the capacitor plates. The potential difference between the "plates" is \(90.0 \mathrm{mV}\). (a) How much energy is stored in this capacitor? (b) How many positive ions are there on the outside of the membrane? Assume that all the ions are singly charged (charge +e).
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A point charge \(q=-2.5 \mathrm{nC}\) is initially at rest adjacent to the negative plate of a capacitor. The charge per unit area on the plates is $4.0 \mu \mathrm{C} / \mathrm{m}^{2}\( and the space between the plates is \)6.0 \mathrm{mm} .$ (a) What is the potential difference between the plates? (b) What is the kinetic energy of the point charge just before it hits the positive plate, assuming no other forces act on it?
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