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Question
For \[r \gg a\], applying the Binomial theorem to \[r_1^2 \approx r^2\left(1 - \frac{2a\cos\theta}{r}\right)\] and retaining only first-order terms, the approximation for \[\frac{1}{r_1}\] is:
Options
\[\frac{1}{r_1} \approx \frac{1}{r}\left(1 - \frac{a\cos\theta}{r}\right)\]
\[\frac{1}{r_1} \approx \frac{1}{r}\left(1 + \frac{2a\cos\theta}{r}\right)\]
\[\frac{1}{r_1} \approx \frac{1}{r^2}\left(1 + a\cos\theta\right)\]
\[\frac{1}{r_1} \approx \frac{1}{r}\left(1 + \frac{a\cos\theta}{r}\right)\]
MCQ
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Solution
Taking the square root gives r₁ ≈ r(1 − a cos θ/r); applying (1 + x)⁻¹ ≈ 1 − x for small x gives 1/r₁ ≈ (1/r)(1 + a cos θ/r). The factor 2a cos θ/r belongs to r₁², and the sign flips upon inversion, so the correction term becomes positive.
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