Gauss’ Law for electric fields The electric field due to a

QUESTION:

Gauss' Law for electric fields The electric field due no a point charge Q is \(\mathbf{E}=\frac{Q}{4 \pi \varepsilon_{0}} \frac{\mathbf{r}}{|\mathbf{r}|^{3}}\), where \(\mathbf{r}=\langle x, y, z\rangle\), and \(\varepsilon_{0}\) is a constant.

a. Show that the flux of the field across a sphere of radius a centered at the origin is \(\iint_{S} \mathbf{E} \cdot \mathbf{n} d S=\frac{Q}{\varepsilon_{0}}\).

b. Let S be the boundary of the region between two spheres centered at the origin of radius a and b with a < b. Use the Divergence Theorem to show that the net outward flux across S is zero.

c. Suppose there is a distribution of charge within a region D. Let q(x, y, z) be the charge density (charge per unit volume). Interpret the statement that \(\iint_{S} \mathbf{E} \cdot \mathbf{n} d S=\frac{1}{\varepsilon_{0}} \iiint_{D} q(x, y, z) d V\)

d. Assuming E satisfies the conditions of the Divergence q Theorem. conclude from pan (c) that \(\nabla \cdot \mathbf{E}=\frac{q}{\varepsilon_{0}}\)

e. Because the electric force is conservative, it has a potential . 7 q function \(\varphi\). From pan (d) conclude that \(\nabla^{2} \varphi=\nabla \cdot \nabla \varphi=\frac{q}{\varepsilon_{0}}\)

Text Transcription:

iint_S E cdot n dS = 1/varepsilon_0 iiint_D q(x, y, z)dV

nabla cdot E = q/ varepsilon_0

nabla^2 varhpi = nabla cdot nabla_varphi = 9/ varepsilon_0

iint_S E cdot n dS = Q/varepsilon_0

E = Q/4pi varepsilon_0 r/|r|^3

r = langle x, y, z rangle