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:

विकल्प

• \[\frac{1}{r_1} \approx \frac{1}{r}\left(1 - \frac{a\cos\theta}{r}\right)\]

• \[\frac{1}{r_1} \approx \frac{1}{r}\left(1 + \frac{2a\cos\theta}{r}\right)\]

• \[\frac{1}{r_1} \approx \frac{1}{r^2}\left(1 + a\cos\theta\right)\]

• \[\frac{1}{r_1} \approx \frac{1}{r}\left(1 + \frac{a\cos\theta}{r}\right)\]

Answer 1

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.