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How could I simplify the following equation (C1) to become the next equation (C2) knowing that $r \gg d$ ($d$ is significantly smaller than $r$)

$$\begin{align*} C_1 &= \frac{\epsilon_0 \pi r^2}{d + r - \sqrt{r^2+d^2}}\\ C_2 &= \frac{\epsilon_0 \pi r^2}{d} \end{align*}$$

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$\sqrt{r^2+d^2}=r+\frac{d^2}{2r}+\cdots$ –  Guess who it is. Nov 8 '11 at 3:19
How do you get to that sequence? –  JFB Nov 8 '11 at 3:48
@JFB: It comes from the Taylor series for the square root.$\sqrt{r^2+d^2}=r\sqrt{1+\frac{d^2}{r^2}}\approx r(1+\frac{d^2}{2r^2})$. The general case is $(1+x)^n \approx 1+nx$ for $x \ll 1$ –  Ross Millikan Nov 8 '11 at 4:07

3 Answers 3

Although you say 'simplify,' it isn't actually an equality that you're looking for, But if r is way bigger than d, than $\sqrt{r^2 + d^2} \approx \sqrt{r^2}$.

Can you finish it from there?

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up vote 1 down vote accepted

Ok, I got it:

$$\begin{align*} (1+x)^n&=&1+nx+\cdots \\ \implies \sqrt{r^2+d^2}&=&\left(r+\frac{d^2}{r}\right)^{1/2} \\ &\approx& r + \frac{d^2}{2r}+\cdots \approx r \end{align*}$$

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To add: you already said $d$ is tiny relative to $r$; thus, $d^2$, $d^3$, and higher powers are even tinier, and dividing by (powers of) $r$ makes them tinier than they already are. –  Guess who it is. Nov 8 '11 at 7:32

Well, if $r \gg d$, then $\sqrt{r^2+d^2} \approx \sqrt{r^2}$. You can probably take it from here.

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