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Chapter 22: Problem 68

Show that the result of Example 22.8 approaches the field of a point charge for \(x \gg a\). (Hint: You'll need to apply the binomial approximation from Appendix A to the expression \(1 / \sqrt{x^{2}+a^{2}}\) )

Short Answer

Expert verified
After applying the binomial approximation to \(1 / \sqrt{x^{2}+a^{2}}\), and substituting it into the field expression gives by Example 22.8, we find that for \(x \gg a\), the expression looks very much like the field of a point charge.

Step by step solution

01

Apply binomial approximation

The binomial approximation from Appendix A is \( (1 + x)^n \approx 1 + nx \) for \( |x| \ll 1 \). To apply it to \(1 / \sqrt{x^{2}+a^{2}}\), first modify the expression to suit the binomial approximation form by dividing both numerator and denominator by \(x^2\). The expression becomes \(1 / \sqrt{1+(a/x)^2}\). Now \(a/x\) is much smaller than 1 for \(x \gg a\), and can be denoted by \(k\) (where \(k = a/x\)). So, the expression becomes \(1 / \sqrt{1+k^{2}}\) .
02

Substitute binomial approximation into the expression

Now, with \(n = -0.5\) and \(x = k^2 = (a/x)^{2}\), apply the binomial approximation to get \((1 + (a/x)^{2})^{-0.5} \approx 1 -0.5 * (a/x)^2\) which simplifies to \(1 - (a^{2}/2x^{2})\).
03

Substituting into Example 22.8

Once we have approximated the expression, substitute it into the field expression given in Example 22.8. Look closely to find the parts of that expression that now simplify due to the new form of the term. Update these parts with the simplified form.
04

Replacing small terms

Now we will see terms that contain \((a/x)\) or \((a/x)^2\), or higher. For \(x \gg a\), these are very small terms and approach zero. Replace them with zero and we find that the simplified expression now matches the field of a point charge.

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Most popular questions from this chapter

You're sizing a new electric transmission line, and you can save money with thinner wire. The potential difference between the line and the ground, \(60 \mathrm{m}\) below, is \(115 \mathrm{kV}\). The field at the wire surface cannot exceed \(25 \%\) of the 3 -MV/m breakdown field in air. Neglecting charges in the ground itself, what minimum wire diameter do you specify? (Hint. You'll have to do a numerical calculation.)

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