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Chapter 23: Problem 19

An uncharged capacitor has parallel plates \(5.0 \mathrm{cm}\) on a side, spaced \(1.2 \mathrm{mm}\) apart. (a) How much work is required to transfer \(7.2 \mu \mathrm{C}\) from one plate to the other? (b) How much work is required to transfer an additional \(7.2 \mu \mathrm{C} ?\)

Short Answer

Expert verified
The work required to transfer \(7.2 \mu \mathrm{C}\) of charge from one plate to the other is \(0.14 \, \mathrm{J}\), and the work required to transfer an additional \(7.2 \mu \mathrm{C}\) of charge is \(0.56 \, \mathrm{J}\).

Step by step solution

01

Find the Capacitance

The formula for the capacitance \(C\) of a parallel plate capacitor is given by \(C = \epsilon_0 \frac{A}{d}\), where \(\epsilon_0\) is the permittivity of free space (\(8.85 × 10^{-12} \, \mathrm{C^2/N \cdot m^2}\)), \(A\) is the area of one of the plates and \(d\) is the separation between the plates. Substituting \(A = (0.05 \, \mathrm{m})^2\) and \(d = 1.2 × 10^{-3} \, \mathrm{m}\) into the equation, we find \(C ≈ 1.85 × 10^{-11} \, \mathrm{F}\).
02

Calculate the First Transfer of Charge

For the first transfer of charge, no voltage yet exists across the capacitor as it is uncharged. Therefore, the work done is equivalent to the energy stored in the capacitor. Hence, we use the formula for energy stored in a capacitor, \(W = 1/2CV^2\). However, since we know the charge instead of the voltage, we first use the formula for charge in a capacitor, \(Q = CV\), to substitute for \(V\) in our energy formula. This gives us \(W_1 = Q^2/(2C)\). Substituting \(Q = 7.2 × 10^{-6} \, \mathrm{C}\) and \(C ≈ 1.85 × 10^{-11} \, \mathrm{F}\), we find \(W_1 ≈ 0.14 \, \mathrm{J}\).
03

Calculate the Second Transfer of Charge

For the second transfer of charge, the work done is due to the existing voltage across the capacitor created by the first transfer. Therefore, we use the formula for energy stored in a capacitor by charge once more, but this time we need to calculate the work done with an existing voltage across the plate. Thus, our formula will be \(W_2 = Q^2/(2C)\), where \(Q\) is now \(2 × 7.2 × 10^{-6} \, \mathrm{C} = 14.4 × 10^{-6} \, \mathrm{C}\). Substituting these values, we find \(W_2 ≈ 0.56 \, \mathrm{J}\)

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Most popular questions from this chapter

Two positive point charges are infinitely far apart. Is it possible, using a finite amount of work, to move them until they're a small distance \(d\) apart?

A conducting sphere of radius \(a\) is surrounded by a concentric spherical shell of radius \(b .\) Both are initially uncharged. How much work does it take to transfer charge from one to the other until they carry charges \(\pm Q ?\)

A capacitor stores \(40 \mathrm{mJ}\) of energy when charged to \(100 \mathrm{V}\) (a) How much would it store at \(25 \mathrm{V} ?\) (b) What's its capacitance?

Your company is still stuck with those 2 - \(\mu\) F capacitors from Problem 50. They turn out to be so cheap that their capacitances are all too low, ranging from \(1.7 \mu \mathrm{F}\) to \(1.9 \mu \mathrm{F}\). A colleague suggests you put variable "trimmer" capacitors in parallel with the cheap capacitors and adjust the combination to precisely \(2.00 \mu \mathrm{F} .\) The available trimmers have variable capacitance from \(25 \mathrm{nF}\) to \(350 \mathrm{nF}\). Will they work?

How does the energy density at a certain distance from a negative point charge compare with the energy density at the same distance from a positive point charge of equal magnitude?
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