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Worksheet 2.5 In-Class Activity

1. True or False.

    If we manage to find the maximum value of \(|f^{(n+1)}(c)|\) for \(c\) in the interval between \(a\) and \(x\text{,}\) then we can find the exact error in the Taylor polynomial approximation.
  • True.

  • Incorrect. While finding the maximum value of \(|f^{(n+1)}(c)|\) allows us to find an upper bound for the error, it does not give the exact error.
  • False.

  • Incorrect. While finding the maximum value of \(|f^{(n+1)}(c)|\) allows us to find an upper bound for the error, it does not give the exact error.

2. Approximating The Electric Field due to an Electric Dipole.

Consider an electric dipole consisting of two charges, \(+q\) and \(-q\text{,}\) separated by a distance \(d\text{.}\) The electric field at a point \(P\) located at a distance \(r\) from the positive charge along the axis of the dipole is given by:
\begin{equation*} E = \frac{1}{4\pi\epsilon_0} \left( \frac{q}{r^2} - \frac{q}{(r+d)^2} \right) \end{equation*}
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Figure 2.23. An electric dipole: charges \(+q\) and \(-q\) separated by a distance \(d\text{,}\) with the field point \(P\) a distance \(r\) from the positive charge along the axis.
Use the Taylor series to approximate the electric field \(E\) at point \(P\) for \(d \ll r\text{.}\) Show that the leading term in the approximation is proportional to \(d/r^3\text{.}\)
Solution.
To approximate the electric field \(E\) at point \(P\) for \(d \ll r\text{,}\) we can use the Taylor series expansion for the function \(f(x) = \frac{1}{(r+x)^2}\) around \(x=0\text{.}\)
We have:
\begin{equation*} f(x) = \frac{1}{(r+x)^2} = \frac{1}{r^2} \left( 1 + \frac{x}{r} \right)^{-2} \end{equation*}
Using the binomial series expansion, we can write:
\begin{equation*} \left( 1 + \frac{x}{r} \right)^{-2} = 1 - 2\frac{x}{r} + 3\frac{x^2}{r^2} - 4\frac{x^3}{r^3} + \cdots \end{equation*}
Substituting \(x=d\) and \(x=0\text{,}\) we get:
\begin{equation*} f(d) = \frac{1}{(r+d)^2} = \frac{1}{r^2} \left( 1 - 2\frac{d}{r} + 3\frac{d^2}{r^2} - 4\frac{d^3}{r^3} + \cdots \right) \end{equation*}
\begin{equation*} f(0) = \frac{1}{r^2} \end{equation*}
Now, we can write the electric field \(E\) as:
\begin{equation*} E = \frac{1}{4\pi\epsilon_0} \left( \frac{q}{r^2} - \frac{q}{(r+d)^2} \right) = \frac{q}{4\pi\epsilon_0} \left( \frac{1}{r^2} - \frac{1}{(r+d)^2} \right) \end{equation*}
Substituting the Taylor series expansion for \(f(d)\) and \(f(0)\text{,}\) we get:
\begin{equation*} E = \frac{q}{4\pi\epsilon_0} \left( \frac{1}{r^2} - \frac{1}{r^2} \left( 1 - 2\frac{d}{r} + 3\frac{d^2}{r^2} - 4\frac{d^3}{r^3} + \cdots \right) \right) \end{equation*}
Simplifying, we have:
\begin{equation*} E = \frac{q}{4\pi\epsilon_0} \left( 2\frac{d}{r^3} - 3\frac{d^2}{r^4} + 4\frac{d^3}{r^5} - \cdots \right) \end{equation*}
Thus, the leading term in the approximation is:
\begin{equation*} E \approx \frac{q}{4\pi\epsilon_0} \cdot 2\frac{d}{r^3} \end{equation*}
which shows that the leading term is proportional to \(d/r^3\text{.}\)
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Figure 2.24. The axial field of a dipole, \(E = \frac{1}{4\pi\epsilon_0} \left(\frac{q}{r^2} - \frac{q}{(r+d)^2}\right)\text{,}\) compared with its leading Taylor term for \(d \ll r\text{.}\) The leading term \(\frac{2qd}{4\pi\epsilon_0 r^3} \propto \frac{d}{r^3}\) overshoots slightly at small \(r\) but converges to the exact field as \(r\) grows.

3. How small must the separation be?

A dipole is often treated as a point dipole by replacing the exact axial field
\begin{equation*} E = \frac{1}{4\pi\epsilon_0}\left( \frac{q}{r^2} - \frac{q}{(r+d)^2} \right) \end{equation*}
with its leading-order Taylor term for \(d \ll r\text{,}\)
\begin{equation*} E \approx \frac{1}{4\pi\epsilon_0}\cdot\frac{2qd}{r^3}. \end{equation*}
Suppose this approximation must agree with the exact field to within a relative error of \(1\%\text{.}\) If the field point \(P\) is \(r = 3.0\ \text{cm}\) from the positive charge, how small must the charge separation \(d\) be for the point-dipole approximation to be valid?
Hint.
The dominant error comes from the next term in the expansion. Using \(E = \dfrac{q}{4\pi\epsilon_0}\left( \dfrac{2d}{r^3} - \dfrac{3d^2}{r^4} + \cdots \right)\text{,}\) form the relative error of the leading term and keep only the largest contribution.
Answer.
\(d \lesssim \dfrac{2}{3}(0.01)\,r \approx 0.020\ \text{cm} = 0.20\ \text{mm}.\)
Solution.
From the Taylor expansion,
\begin{equation*} E = \frac{q}{4\pi\epsilon_0}\left( \frac{2d}{r^3} - \frac{3d^2}{r^4} + \cdots \right), \end{equation*}
the leading term is \(E_{\text{lead}} = \dfrac{q}{4\pi\epsilon_0}\cdot \dfrac{2d}{r^3}\text{,}\) and the first neglected term is \(\dfrac{q}{4\pi\epsilon_0}\cdot\dfrac{3d^2}{r^4}\text{.}\) The relative error of the approximation is therefore
\begin{equation*} \frac{\left| E_{\text{lead}} - E \right|}{E} \approx \frac{\dfrac{3d^2}{r^4}}{\dfrac{2d}{r^3}} = \frac{3}{2}\cdot\frac{d}{r}. \end{equation*}
Notice that \(q\) and \(\epsilon_0\) cancel, so the relative error depends only on the ratio \(d/r\text{.}\) Requiring this to be at most \(1\% = 0.01\) gives
\begin{equation*} \frac{3}{2}\cdot\frac{d}{r} \le 0.01 \qquad\Longrightarrow\qquad d \le \frac{2}{3}(0.01)\,r. \end{equation*}
With \(r = 3.0\ \text{cm}\text{,}\)
\begin{equation*} d \le \frac{2}{3}(0.01)(3.0\ \text{cm}) = 0.020\ \text{cm} = 0.20\ \text{mm}. \end{equation*}
So the point-dipole approximation is accurate to \(1\%\) only when the charges are separated by less than about \(0.2\ \text{mm}\) at this distanceβ€”consistent with FigureΒ 2.24, where the two curves visibly merge as \(r\) grows relative to \(d\text{.}\)
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