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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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