Question

Show that for x≫R, equation Ex=σ2ϵ0[1−1(R2/x2+1)√] becomes E=Q/4πϵ0x2, where Q is the total charge on the disk.

This essay question confuses me and any help with an explanation would be very appreciated.

Constants Part C A uniformly charged disk has radius 2.50 cm and carries a total charge of 2.0x10-12 C (Figure 1) Show that for z >R, equation E,- 1 becomes E-Q /4 περζ2, where Q is the total charge on the disk. Figure 1 of 1 Essay answers are limited to about 500 words (3800 characters maximum, including spaces) 3800 Character(s) remaining Submit Request Answer

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Answer 1

Ans.

$\text{Consider the second term of the expression.} \\ \frac{1}{\sqrt{\frac{R^2}{x^2}+ 1}} = \left (1 + \frac{R^2}{x^2}\right )^{-\frac{1}{2}} \\ x \gg R , \text{ so } \frac{R^2}{x^2} \ll 1 \\ \text{Now, using Binomial series} \\ \\ \left (1 + \frac{R^2}{x^2}\right )^{-\frac{1}{2}} =\left(1 + \left( \frac{-1}{2} \right) \frac{R^2}{x^2} \right ) \text{ ignoring other terms as } x\gg R \\ \\ \\ = \left (1- \frac{R^2}{2x^2} \right) \\ \\ \text{Now plugging this into the equation of E} \\ E = \frac{ \sigma }{ 2\epsilon _o}\left(1 -\left (1- \frac{R^2}{2x^2} \right) \right ) \\ \\ = \frac{ \sigma }{ 2\epsilon _o}\left(\frac{R^2}{2x^2} \right ) \\ \text{ Now using }\sigma = \frac{Q}{\pi R^2} \\ \\ \text{Putting this into E} \\ E = \frac{Q}{\pi R^2}\frac{1}{2\epsilon_o}\left(\frac{R^2}{2x^2} \right ) \\ \\ E = \frac{Q}{4 \pi \epsilon_o x^2}$