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Some answers

Griffiths (2.3)
$\frac{\lambda}{4\pi\epsilon_0}[(L/z\sqrt{L^2+z^2}){\mathbf k}
-(1/z\sqrt{L^2+z^2}-1/2z){\mathbf i}]$
Griffiths (2.5)
$\sigma R z/2\epsilon_0(R^2+z^2)^{3/2}$
Griffiths (2.6)
$\frac{\sigma}{2\epsilon_0}[1 - z/\sqrt{R^2+z^2}] $
Griffiths (2.9)
(a) $\rho(r) = 5k\epsilon_0 r^2$;(b) $4\pi k \epsilon_0 R^5$
Griffiths (2.15)
$\mathbf{E} = 0$ if $r < a$. $\mathbf{E} = \hat{\mathbf{r}}(k/\epsilon_0)(r-a)/r^2$ if $a< r < b$. $\mathbf{E} = \hat{\mathbf{r}}(k/\epsilon_0)(b-a)/r^2$ if $b< r $. $\vert\mathbf{E}\vert$ is increasing at $r=a$, has a maximum at $r = 2a$ if $b > 2a$.
Griffiths (2.18)
$(\rho/3\epsilon_0)\mathbf{d}$
Griffiths (2.26)
$(\rho h/2\epsilon_0)(\log(\sqrt{2}+1)-1)$
Jackson (1.3)
(a) $(1/4\pi R^2)\delta(r-R)$;(b) $(1/2\pi R)\delta(r-b)$ (c) $(Q/\pi R^2)\delta(z)\Theta(R-r)$ (d) $(Q/\pi R^2 r)\delta(\theta-\pi/2)\Theta(R-r)$
Jackson (1.4)
$q\delta(\mathbf{r})-(q\alpha^3/8\pi)e^{-\alpha r}$



Charudatt Kadolkar 2007-01-30