Let [itex]\;\vec r\;'\; [/itex] be position vector pointing to the source. [itex]\;\vec r\;[/itex] be the position vector pointing to a field point. Therefore the distance from the source and field point is:(adsbygoogle = window.adsbygoogle || []).push({});

[tex] \vec {\eta} =\vec r -\vec r\;' \;\hbox { and }\; \eta = \sqrt{r^2+r'^2 - 2rr'cos\theta}= r \sqrt{1+\left (\frac{r'}{r}\right )^2 -2\frac{r'}{r} cos \theta}[/tex]

Where [itex]\;\theta\;[/itex] is the angle between the two vector.

For r'<<r, Why is the book than say

[tex] \vec {\eta} =\vec r -\vec r\;' \;\hbox { and }\; \eta = \sqrt{r^2+r'^2 - 2rr'cos\theta}= r \sqrt{1+\left (\frac{r'}{r}\right )^2 -2\frac{r'}{r} cos \theta} = r \left (1-\frac {r'}{r} cos \theta\right )[/tex]

I know for r'<<r, [itex]\left ( \frac {r'}{r}\right )^2\approx \;0 [/itex]. But how can you remove the square root.

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# How do I derive the step on the formula of distance of source to field point.

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