hw5_solutions - TIlE STATIC ELECTRIC FIELD 101 where the...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
TIlE STATIC ELECTRIC FIELD 101 where the last two terms are the so-called Exponential Integrals for which there is no closed form solution but which are well tabulated t . (d) The result can be shown by simple substitution into the cylindrical coordinate version of Poisson's equation, namely 1 8 2 C1> ~CI> r 2 8cjJ2 + 8z 2 ""--" ~ No variation in </J No variation in z 2 I 8 ( 8C1» V CI>=-- r- + r8r 8r -p(r) =-- Note that differentiation of the Exponential Integral terms simply yield the integrands. The electron charge density in a hydrogen atom. (a) Since there is spherical symmetry, we consider a spherical Gaussian surface S with radius r and apply Gauss's law as where where we used the integral which can easily be shown using integration by parts twice. As a result, the electric field due to the electron cloud only can be found as E 1 4qe [a 3 -2r/a (ar 2 a 2 r a 3 )] - --e -+ +- r 41T1:or2 a 3 4 2 2 4 - 41T~:r2 { 1- e- 2r / a [2(~r + 2 (~) + I] } Note that the constant a in the above expression is called the Bohr radius given by a ~ 0.529 x 10- 10 m. The electric potential can be found by integrating the electric field. We find i r qe -2r/a[1 1] qe 1 CI>(r) = - Er(r)dr = -- e - + - - --- 00 41TfO a r 41TfO r . See Chapter 5 of M. Abramowitz and I. A. Stegun, Handbook ofMathematical Functions With Formulas, .. aphs, and mathematical Tables, National Bureau of Standards, Tenth Printing, 1972.
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
102 niB STATIC ELECTRIC FIELD (b) Using superposition principle, the total electric field is given by where ETc is the field due to the electron cloud only (found in part (a» and E1'p is the field due to the proton charge -qe located at the origin given by The electric potential can be found by integrating the electric field. We find 4·22. Spherical sheD ofcharge. (a) The total charge in the spherical shell region specified by a S r S b is given by Q = [ p(r')dv' = [21f r [b p(r')r,2 sin fJ'dr'dfJ'drp' iv' io io ia b K lb =(21T)(2) 2r,2dr, = 41TK dr' = 41TK(b - a) l a r' a (b) Due to spherical symmetry, the electric field has the form E = t E1'(r). We consider a spherical Gaussian surface with radius r and apply Gauss's law:
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 11/16/2010 for the course ECE 305 taught by Professor Staff during the Summer '08 term at Michigan State University.

Page1 / 6

hw5_solutions - TIlE STATIC ELECTRIC FIELD 101 where the...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online