New intregral same old problem

[Phymath]

A stright line segment of length 2L is placed on the z-axis with its midpoint at the origin. The segment has a linear charge density given by...
$$\lambda = \frac{aQ}{z^2 + a^2}$$
where Q and a are constants, a>0 find the electrostatic potential of this charge distribution at a point on the x-axis in Cartesian coords.

$$\Phi = \int_{-L}^{L} \frac{k_e dt \frac{aQ}{t^2+a^2}}{\sqrt{x^2 + t^2}}$$
$$= k_e Q a \int_{-L}^{L} \frac{dt}{(t^2 +a^2) \sqrt{t^2 + x^2}}$$

what do now?...no idea what subsitution will make this work,any suggestions?

 

[dextercioby]

Yes,u may put your integral under the form
$$\Phi=2kQa\int_{0}^{L}\frac{dt}{(t^{2}+a^{2})x\sqrt{\left(\frac{t}{x}\right) ^{2}+1}}$$

and do the substitution:
$$\frac{t}{x}=\sinh u$$

Daniel.

 

[dextercioby]

Okay,if you do that substitution,u'll end up with these integrals (actually the second).

Daniel.

 

[Phymath]

ya...instead of that good idea (which it is) how about...
$$= k_e Q a \int_{-L}^{L} \frac{dt}{(t^2 +a^2) \sqrt{t^2 + x^2}}$$

$$= \frac{k_e Q a}{x} \int_{-L}^{L} \frac{dt}{(t^2 +a^2) \sqrt{(\frac{t}{x})^2 + 1}}, \ \frac{t}{x} = tan(u)$$

$$t^2 = x^2 tan^2 (u), \ dt = x sec^2 (u) du$$

$$= \frac{k_e Q a}{x} \int_{tan^{-1}(-L/x)}^{tan^{-1}(L/x)} \frac{dt}{(x^2 tan^2 (u) +a^2) \sqrt{tan^2 u + 1}}$$

$$= \frac{k_e Q a}{x} \int_{tan^{-1}(-L/x)}^{tan^{-1}(L/x)} \frac{x sec^2 (u) du}{(x^2 tan^2 (u) +a^2) sec(u)}$$

$$= k_e Q a \int_{tan^{-1}(-L/x)}^{tan^{-1}(L/x)} \frac{sec (u) du}{(x^2 tan^2 (u) +a^2)}, \ c= \frac{a}{x}$$

$$= k_e Q a \int_{tan^{-1}(-L/x)}^{tan^{-1}(L/x)} \frac{sec (u) du}{x^2( tan^2 (u) + c^2)}$$

$$= \frac{k_e Q a}{x^2} \int_{tan^{-1}(-L/x)}^{tan^{-1}(L/x)} \frac{sec (u)}{(tan^2 (u) + (\frac{a}{x})^2}$$


$$=\frac{2 tan^{-1}(\frac{L\sqrt{x^2-a^2}}{a\sqrt{x^2+L^2}}{1}$$

does this all look legal? if there's an x^2 in the last one, idk why its there it's not supposed to be...

 

[dextercioby]

Correct that latex errors.It looks good.That substitution leads to the same answer.I still prefer hyperbolic trig.functions.

Anyway,your problem is solvable.It depends on u which eq.u prefer most.

Daniel.