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Chapter 23: Problem 37

A charge \(Q_{0}\) is at the origin. A second charge, \(Q_{x}=2 Q_{0},\) is brought from infinity to the point \(x=a, y=0 .\) Then a third charge \(Q_{y}\) is brought from infinity to \(x=0, y=a .\) If it takes twice as much work to bring in \(Q_{y}\) as it did \(Q_{x},\) what's \(Q_{y}\) in terms of \(Q_{0} ?\)

Short Answer

Expert verified
The charge \(Q_{y}\) is \(4/3\) times the charge \(Q_{0}\), i.e., \(Q_{y} = 4/3 * Q_{0}\)

Step by step solution

01

Calculate work done to bring the second charge.

To compute the work done to bring the charge \(Q_{x}\), we use: \(W_{Q_{x}} = k * Q_{0} * Q_{x} / a\). Substituting \(Q_{x}\) as \(2Q_{0}\), we get: \(W_{Q_{x}} = 2k * Q_{0}^2 / a\)
02

Calculate work done to bring the third charge.

To compute the work done to bring charge \(Q_{y}\), we must consider that it experiences the electric field due to both \(Q_{0}\) and \(Q_{x}\). Hence the work done \(W_{Q_{y}}\) will be: \(W_{Q_{y}} = k * Q_{0} * Q_{y} / a + k * Q_{x} * Q_{y} / a = k * Q_{y} * (Q_{0} + 2*Q_{0}) / a = 3k * Q_{0} * Q_{y} / a\)
03

Equate the work done equations and solve for \(Q_{y}\)

Given \(W_{Q_{y}} = 2W_{Q_{x}}\), we can replace \(W_{Q_{y}}\) and \(W_{Q_{x}}\) in terms of \(Q_{0}\) and \(Q_{y}\) from steps 1 and 2: 3k * Q_{0} * Q_{y} / a = 2 * 2k * Q_{0}^2 / a After simplification, we obtain: \(Q_{y} = 4/3 * Q_{0}\)

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