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Chapter 23: Problem 13

Four \(50-\mu \mathrm{C}\) charges, initially far apart, are brought onto a line where they're spaced at 2.0 -cm intervals. How much work does it take to assemble this charge distribution?

Short Answer

Expert verified
The total work done in assembling this charge distribution is given by the sum of potential energies, \(W = U_{2} + U_{3}\), calculated using the formula for potential energy between two charges.

Step by step solution

01

Understand the Fundamentals

Firstly, remember that the formula for the potential energy U between two charges is given as : \[U = k \cdot \frac{q1 \cdot q2}{r}\] where k is Coulomb’s constant (\(9 x 10^9 Nm^2/C^2\)), \(q1\) and \(q2\) are the charges and r is the distance between them. Note that potential energy is a scalar, so we simply add up the energies.
02

Work Needed to Bring the First and Fourth Charge

The work needed to bring the first and fourth charge from infinity to its current position is 0, because there is no charge at the beginning.
03

Work Needed to Bring the Second Charge

The work to bring the second charge is done against the force exerted by the first charge. So it can be calculated as: \[U_{2} = k \cdot \frac{q \cdot q}{r} = 9 \times 10^9 \cdot \frac{50 \times 10^{-6} \cdot 50 \times 10^{-6}}{0.02}\]
04

Work Needed to Bring the Third Charge

The work to bring the third charge is done against the force exerted by the first and second charge. The distance between third charge and the first charge is 2r and between third and the second charge is r. Due to the scalar nature of potential energy, we will add the energies required: \[U_{3} = k \cdot \frac{q \cdot q}{2r} + k \cdot \frac{q \cdot q}{r} = \frac{3}{2} \times 9 \times 10^9 \cdot \frac{50 \times 10^{-6} \cdot 50 \times 10^{-6}}{0.02}\]
05

Total Work Required

The total work done in assembling the charge distribution is simply the sum of the work done to bring each charge individually: \(W = U_{2} + U_{3}\)

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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?

Find the capacitance of a capacitor that stores \(350 \mu \mathrm{J}\) when the potential difference across its plates is \(100 \mathrm{V}\).

A camera requires \(5.0 \mathrm{J}\) of energy for a flash lasting \(1.0 \mathrm{ms}\). (a) What power does the flashtube use while it's flashing? (b) If the flashtube operates at \(200 \mathrm{V},\) what size capacitor is needed to supply the flash energy? (c) If the flashtube is fired once every \(10 \mathrm{s},\) what's its average power consumption?

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?

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 \(7.2 \mu \mathrm{C} ?\)
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