Gauss' Law for electric fields The electric field due to a pointQ r4TE, |r|3"charge Q is Ewhere r = (x, y, z), and ɛ, is aconstant.a. Show that the flux of the field across a sphere of radius a cen-tered at the origin is ,E -n dS = º.b. Let S be the boundary of the region between two spheres cen-tered at the origin of radius a and b, respectively, with a < b.Use the Divergence Theorem to show that the net outward fluxacross 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||E·n ds =- q(x, y, z) dV.d. Assuming E satisfies the conditions of the DivergenceTheorem on D, conclude from part (c) that V · E = 4.e. Because the electric force is conservative, it has a potentialfunction o. From part (d), conclude that v²p = V • Vọ =03

Step 1: Given: The electric field due to a point charge is given by : E = Q4πεorr3 where r = x,y,z…

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Question

Gauss' Law for electric fields The electric field due to a point
Q r
4TE, |r|3"
charge Q is E
where r = (x, y, z), and ɛ, is a
constant.
a. Show that the flux of the field across a sphere of radius a cen-
tered at the origin is ,E -n dS = º.
b. Let S be the boundary of the region between two spheres cen-
tered at the origin of radius a and b, respectively, 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
||
E·n ds =- q(x, y, z) dV.
d. Assuming E satisfies the conditions of the Divergence
Theorem on D, conclude from part (c) that V · E = 4.
e. Because the electric force is conservative, it has a potential
function o. From part (d), conclude that v²p = V • Vọ =
03

Expert Solution

Step 1

Step 1:

Given: The electric field due to a point charge is given by : $E = \frac{Q}{4 {πε}_{o}} \frac{r}{{|r|}^{3}} w h e r e r = <x, y, z>$

Divergence Theorem : $\underset{V}{\int \int \int} \nabla F d V = \iint_{S} F . n d S$

Step 2

Step 2:

$\iint_{s} E . n d S, w e k n o w n = \frac{r}{| r |} . \iint_{s} E . n d S = \iint_{s} \frac{Q}{4 {πε}_{o}} \frac{r}{{|r|}^{3}} . \frac{r}{| r |} d S \iint_{s} E . n d S = \frac{Q}{4 {πε}_{o}} \iint_{s} \frac{{|r|}^{2}}{{|r|}^{4}} d S \iint_{s} E . n d S = \frac{Q}{4 {πε}_{o}} \iint_{s} \frac{1}{{|r|}^{2}} d S N o t e : R a d i u s o f s p h e r e i s a . P u t |r| = a w e g e t, \iint_{s} E . n d S = \frac{Q}{4 {πε}_{o}} \iint_{s} \frac{1}{a^{2}} d S \iint_{s} E . n d S = \frac{Q}{4 {πε}_{o}} \frac{1}{a^{2}} \iint_{s} d S \iint_{s} E . n d S = \frac{Q}{4 {πε}_{o}} \frac{1}{a^{2}} 4 {πa}^{2} \iint_{s} E . n d S = \frac{Q}{ε_{o}}$

Step 3

Step 3:

According to Divergence theorem,

$\underset{V}{\int \int \int} \nabla F d V = \iint_{S} F . n d S = \iint_{S_{1}} F . n_{2} d S - \iint_{S_{2}} F . n_{1} d S W e k n o w t h a t d i v e r g e n c e o f r a d i a l f i e l d F = \frac{r}{{| r |}^{p}} = \frac{3 - p}{{| r |}^{p}} T h e r e f o r e, \nabla \cdot F = \frac{3 - p}{{| r |}^{p}} \nabla \cdot F = \frac{3 - 3}{{| r |}^{p}} f o r p = 3 \nabla \cdot F = \frac{3 - 3}{{| r |}^{3}} \nabla \cdot F = 0 N o w w e w a n t t o f i n d d i v e r g e n c e o f g i v e n e l e c t r i c f i e l d E = \frac{Q}{4 {πε}_{o}} \frac{r}{{|r|}^{3}} w h e r e r = <x, y, z> \underset{V}{\int \int \int} \nabla \cdot E d V = \iint_{S} E . n d S \nabla \cdot E = \nabla \cdot \frac{Q}{4 π ε_{o}} \frac{r}{{| r |}^{3}} \nabla \cdot E = \frac{Q}{4 π ε_{o}} \nabla \cdot \frac{r}{{| r |}^{3}} \nabla \cdot E = 0 T h e r e f o r e, \underset{D}{\int \int \int} \nabla \cdot E d V = \underset{D}{\int \int \int} 0 . d V = 0$

Therefore, Net outward flux across the surface S is zero.

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