Question

Problem 1. (30 points) Consider a line with uniform charge density io along a z-axis that extends from in the diagram below a to- u, as seen 2a As found in the previous exam, the algebraic form of the electric potential along the z axis is (a) we want to expand this electric potential function as an infinite series for the case when a. However, the logarithm function is not amenable to a binomial expansion, so we have to use a different method. In the CRC Mathematical Tables book, you can find a logarithmic series expansion for the function loge 1-x when x < 1, where the natural logarithm, In as we write it, is written as log, by the CRC reference book. The infinite series for this function can be found in the "Series Expansions" section of the "Calculus" chapter. Write down the equality between the function above and its infinite series and cite page number and edition of the CRC. Show how to define x as a ratio of z and a, with z> a, in order that the condition x < 1 is fulfilled Then show algebraically that zta 1-x, where x is replaced by your ratio ofz and a. (b) Use the series expansion you identified in part (a) to expand the electric potential given above as an infinite series whenz> a, explicitly writing out the first three terms in the series. Note that V- Va), so this expansion should be written as powers of z, not as powers of z. Now begins the work to determine the electric potential ir,0) everywhere in the region of spacer> a where r and θ are the usual spherical coordinates.

ANSWER ALL PARTS

Answer 1

(a)

(a) ln 1 + x/1 - x = 2(x + x^3/3 + x^5/5 + ........) = 2 sigma^infinity_n = 0 x^2n + 1/2n + 1 x = a/z z + a/z - a = z(1 + a/z)/z(1 - a/z) = 1 + a/z/1 - a/z for x = a/z z + a/z - a = z(1 + a/z)/z(1 - a/z) = 1 + a/z/1 - a/z = 1 + x/1 - x (b) V(z) = lambda_0/4 pi elementof_0 ln 1 + x/1 - x = lambda_0/2 pi elementof_0 (x + x^3/3 + x^5/5 + ......) since x = a/z V(z) = lambda_0/4 pi elementof_0 ln 1 + x/1 - x = lambda_0/2 pi elementof_0 (a/z + a^3/3z^3 + a^5/5z^5 + .....) (c) the general potential equation is V(r, theta) = sigma^infinity_l = 0 (A_l r^l + B_l/r^l +) P_l (cos theta) as r rightarrow infinity implies r^l rightarrow infinity then the implies A_l r^l rightarrow infinity V(r, theta) = infinity + sigma^infinity_l = 0 B_l/r^l + 1

$\frac{z+a}{z-a} = \frac{z(1+a/z)}{z(1-a/z)} = \frac{1+a/z}{1-a/z}$

for x = a/z

$\Rightarrow \frac{z+a}{z-a} = \frac{z(1+a/z)}{z(1-a/z)} = \frac{1+a/z}{1-a/z} = \frac{1+x}{1-x}$

(b)

$V(z) = \frac{\lambda_{0}}{4\pi \epsilon_{0}} ln\frac{1+x}{1-x} = \frac{\lambda_{0}}{2\pi \epsilon_{0}}\left ( x + \frac{x^{3}}{3} + \frac{x^{5}}{5} + ... \right )$

since x = a/z

$\Rightarrow V(z) = \frac{\lambda_{0}}{4\pi \epsilon_{0}} ln\frac{1+x}{1-x} = \frac{\lambda_{0}}{2\pi \epsilon_{0}}\left ( \frac{a}{z} + \frac{a^{3}}{3z^{3}} + \frac{a^{5}}{5z^{5}} + ... \right )$

(c) the general potential equation is

$V(r,\theta) = \sum_{l= 0}^{\infty}(A_{l}r^{l} + \frac{B_{l}}{r^{l+1}})P_{l}(cos\theta)$

as $r\rightarrow \infty \Rightarrow r^{l}\rightarrow \infty$

then the $\Rightarrow A_{l}r^{l}\rightarrow \infty$

$\Rightarrow V(r,\theta) = \infty + \sum_{l = 0}^{\infty}\frac{B_{l}}{r^{l+1}}P_{l}(cos\theta)$ but since potential must go to zero as r goes to infinity, implies $A_{l}r^{l} = 0 \Rightarrow A_{l} = 0$ for all l

(d)

$\Rightarrow V(r,\theta) = \sum_{l = 0}^{\infty}\frac{B_{l}}{r^{l+1}}P_{l}(cos\theta)$

on z axis $\theta = 0 \Rightarrow cos\theta = cos 0 = 1$

and

Thus on z axis

$V(r,\theta) = V(z) = \sum_{l=0}^{\infty} \frac{B_{l}}{z^{l+1}}$

$V(z) =\frac{B_{0}}{z} + \frac{B_{1}}{z^{2}} + \frac{B_{2}}{z^{3}} + ...$

in part (b) we found that

$\Rightarrow V(z) = \frac{\lambda_{0}}{4\pi \epsilon_{0}} ln\frac{1+x}{1-x} = \frac{\lambda_{0}}{2\pi \epsilon_{0}}\left ( \frac{a}{z} + \frac{a^{3}}{3z^{3}} + \frac{a^{5}}{5z^{5}} + ... \right )$

thus B1 = B3 = B5 = ... = 0

and $B_{0} = \frac{\lambda_{0}a}{2\pi \epsilon_{0}}$ , $B_{2} = \frac{\lambda_{0}a^{3}}{6\pi \epsilon_{0}}$ and $B_{4} = \frac{\lambda_{0}a^{5}}{10\pi \epsilon_{0}}$ are the first three non-zero coefficients