/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 83 A parallel plate capacitor has a... [FREE SOLUTION] | 91影视

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Chapter 17: Problem 83

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?

Short Answer

Expert verified
Answer: The work done by the external agent to double the plate separation while keeping the charge constant is \(5.3 \times 10^{-5}\ \mathrm{J}\).

Step by step solution

01

Write down the given information

We are given: - Initial capacitance of the capacitor, \(C_1 = 1.20 \mathrm{nF} = 1.20 \times 10^{-9} \mathrm{F}\) - Charge on each plate, \(Q = 0.80 \mu \mathrm{C} = 0.80 \times 10^{-6} \mathrm{C}\) - Work done by an external agent, \(W = ?\)
02

Calculate Initial Voltage

Find the initial voltage across the capacitor using the formula \(V_1 = \frac{Q}{C_1}\): \(V_1 = \frac{0.80 \times 10^{-6}}{1.20 \times 10^{-9}} = \frac{2}{3} \times 10^3\ \mathrm{V}\)
03

Calculate Initial Energy

Find the initial energy stored in the capacitor using the formula \(U_1 = \frac{1}{2}C_1V_1^2\): \(U_1 = \frac{1}{2}(1.20 \times 10^{-9})\left(\frac{2}{3} \times 10^3\right)^2 = \frac{1}{2}(1.20 \times 10^{-9})(\frac{4}{9} \times 10^6)\) \(U_1 = \frac{1}{2}(1.20 \times \frac{4}{9}) = \frac{8}{15} \times 10^{-4}\ \mathrm{J}\)
04

Calculate Final Capacitance

When the distance between the plates is doubled, the capacitance is halved. Therefore, the final capacitance is: \(C_2 = \frac{C_1}{2} = \frac{1.20 \times 10^{-9}}{2} = 0.60 \times 10^{-9} \mathrm{F}\)
05

Calculate Final Voltage

Calculate the final voltage using the formula \(V_2 = \frac{Q}{C_2}\): \(V_2 = \frac{0.80 \times 10^{-6}}{0.60 \times 10^{-9}} = \frac{4}{3} \times 10^3\ \mathrm{V}\)
06

Calculate Final Energy

Find the final energy stored in the capacitor using the formula \(U_2 = \frac{1}{2}C_2V_2^2\): \(U_2 = \frac{1}{2}(0.60 \times 10^{-9})\left(\frac{4}{3} \times 10^3\right)^2 = \frac{1}{2}(0.60 \times 10^{-9})(\frac{16}{9} \times 10^6)\) \(U_2 = \frac{1}{2}(0.60 \times \frac{16}{9}) = \frac{32}{30} \times 10^{-4}\ \mathrm{J}\)
07

Calculate Work Done

Find the work done by the external agent to double the plate separation by subtracting the initial energy from the final energy: \(W = U_2 - U_1 = \left(\frac{32}{30} - \frac{8}{15}\right) \times 10^{-4}\) \(W = \frac{16}{30} \times 10^{-4} = 0.053 \times 10^{-4}\ \mathrm{J}\) The work done by the external agent to double the plate separation while keeping the charge constant is \(5.3 \times 10^{-5}\ \mathrm{J}\).

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Most popular questions from this chapter

In \(1911,\) Ernest Rutherford discovered the nucleus of the atom by observing the scattering of helium nuclei from gold nuclei. If a helium nucleus with a mass of \(6.68 \times 10^{-27} \mathrm{kg},\) a charge of \(+2 e,\) and an initial velocity of \(1.50 \times 10^{7} \mathrm{m} / \mathrm{s}\) is projected head-on toward a gold nucleus with a charge of \(+79 e\), how close will the helium atom come to the gold nucleus before it stops and turns around? (Assume the gold nucleus is held in place by other gold atoms and does not move.)
An electron is accelerated from rest through a potential difference $\Delta V\(. If the electron reaches a speed of \)7.26 \times 10^{6} \mathrm{m} / \mathrm{s},$ what is the potential difference? Be sure to include the correct sign. (Does the electron move through an increase or a decrease in potential?)
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A parallel plate capacitor has a capacitance of \(2.0 \mu \mathrm{F}\) and plate separation of \(1.0 \mathrm{mm} .\) (a) How much potential difference can be placed across the capacitor before dielectric breakdown of air occurs \(\left(E_{\max }=3 \times 10^{6} \mathrm{V} / \mathrm{m}\right) ?\) (b) What is the magnitude of the greatest charge the capacitor can store before breakdown?

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