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Chapter 21: Problem 53

A negative charge, \(-q\), is fixed at the coordinate (0,0) It is exerting an attractive force on a positive charge, \(+q,\) that is initially at coordinate \((x, 0)\). As a result, the positive charge accelerates toward the negative charge. Use the binomial expansion \((1+x)^{n} \approx 1+n x,\) for \(x \ll 1,\) to show that when the positive charge moves a distance \(\delta \ll x\) closer to the negative charge, the force that the negative charge exerts on it increases by \(\Delta F=2 k q^{2} \delta / x^{3}\) .

Short Answer

Expert verified
Answer: The approximate change in force exerted by the fixed negative charge on the positive charge after the positive charge moves a small distance closer to the negative charge is given by the expression: \(\Delta F = \frac{2kq^2\delta}{x^3}\).

Step by step solution

01

Write down the expression for Coulomb's force

The force between two point charges can be calculated using Coulomb's law: \(F = k\frac{q_1q_2}{r^2}\). In this case, \(q_1 = -q\), \(q_2 = +q\), and the distance between the charges is \(r\). The charges have opposite signs, resulting in an attractive force.
02

Determine the initial force, \(F_{initial}\), between the charges

Initially, the positive charge is at coordinate \((x, 0)\), and the distance between the two charges is \(r = x\). The initial force, \(F_{initial}\), between the charges can be calculated using the formula from Step 1: \(F_{initial} = k\frac{-q^2}{x^2}\).
03

Determine the final force, \(F_{final}\), after moving the positive charge a distance \(\delta\) closer to the negative charge

The positive charge moves \(\delta\) closer to the negative charge, so its new coordinate is \((x-\delta,0)\). Thus, the new distance between the charges is \(r = x-\delta\). The final force, \(F_{final}\), can also be calculated using the formula from Step 1: \(F_{final} = k\frac{-q^2}{(x-\delta)^2}\).
04

Use the binomial expansion to approximate \(F_{final}\)

Since \(\delta \ll x\), \((x-\delta) \approx x\left(1 - \frac{\delta}{x}\right)\). Using the binomial expansion with \(n=-2\) and \(x = -\frac{\delta}{x}\), we get: \(F_{final} = k\frac{-q^2}{x^2(1-\frac{\delta}{x})^{-2}} \approx k\frac{-q^2}{x^2(1+2\frac{\delta}{x})}\)
05

Calculate the change in force, \(\Delta F\)

The change in force, \(\Delta F\), is given by the difference between the final and initial forces: \(\Delta F = F_{final} - F_{initial}\). Substituting the expressions for \(F_{final}\) and \(F_{initial}\) from Steps 2 and 4, we get: \(\Delta F = k\frac{-q^2}{x^2(1+2\frac{\delta}{x})} - k\frac{-q^2}{x^2} = k\frac{q^2}{x^2}\left(1-\frac{1}{1+2\frac{\delta}{x}}\right) = k\frac{q^2}{x^2}\left(\frac{2\frac{\delta}{x}}{1+2\frac{\delta}{x}}\right)\) Since \(\delta \ll x\), we can approximate the denominator as \(1+2\frac{\delta}{x} \approx 1\): \(\Delta F = k\frac{q^2}{x^2}\left(2\frac{\delta}{x}\right) = \frac{2kq^2\delta}{x^3}\)

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