Question

2. Often, formulas describing phenomena in the sciences are given in terms of many variables many of which are merely constants. In the realm of electromagnetist, an electrically charged wire creates an "electric field" at a distance D from the wire that we have mered. The component El of the field perpendicular to the wire (in Newtons per Coulomb) is D dar (12 + D23/2 Compute the integral. (You should treat D as you would treat a number, as if you measured it.)

Answer 1

ANSWER:-

Obten,formulasdescibing phenomena in the science are given in terms of many variables, many of which are merely constants.

Given that:-

PI2 El D dar (x2 + D2)] 1'1

$\large E_\perp=\int _{\theta_1}^{\theta_2}\frac{D}{(D^2Tan^2\theta+D^2)^\frac{3}{2}}Dsec^2\theta d\theta$

Put:-

$\large x=D\, \, tan\, \, \theta$

$\large dx=D\, \, sec^{2}\, \, \theta \, \, d\theta$

$\large For x=x_{1}\Rightarrow \theta=\theta_{1}=Tan^{-1}\left ( \frac{x_{1}}{D} \right )$

$\large For x=x_{2}\Rightarrow \theta=\theta_{2}=Tan^{-1}\left ( \frac{x_{2}}{D} \right )$

$\large =\int_{\theta_{1 }}^{ \theta_{2} }\frac{D^{2}sec^{2}\theta }{[D^{2}(1+tan^{2}\theta)]^{3/2}}d \theta$

$\large =\int_{\theta_{1 }}^{ \theta_{2} }\frac{D^{2}sec^{2}\theta }{( D^{2}sec^{2}\theta)^{3/2}}d \theta$

$\large =\int_{\theta_{1 }}^{ \theta_{2} }\frac{D^{2}sec^{2}\theta }{( D^{3}sec^{3}\theta)}d \theta$

$\large =\int_{ \theta_{1}}^{ \theta_{2} }\frac{1}{D sec \theta }d \theta =\frac{1}{D}\int_{ \theta_{1}}^{ \theta_{2} }cos \theta d \theta$

$\large = \frac{1}{D} \left ( sin \theta\right )^{\theta_{2} }_{ \theta_{1} }$

$\large = \frac{1}{D} \left ( sin \theta_{2}-sin \theta_{1}\right )$

$\large \theta _{1}=tan^{-1}\left ( \frac{x_{1}}{D} \right )$

$\large \theta _{2}=tan^{-1}\left ( \frac{x_{2}}{D} \right )$