Electric Flux and Gauss\' Law - Key PDF

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Electric Flux and Gauss' Law - Key 1.

The electric flux through a cubical box 8 cm on a side is 1.2 x 103 Nm2/C. What is the total charge enclosed by the box?

 1.2 x103 V  0.08m 

Nm C

2

3

qin 0 qin  0  

2  C2   3 Nm  qin  8.85 x10  12 x 1.2 10  2  Nm   C  

qin 1.06 x10  8 C

A total charge of 5 x 10-8 C is distributed uniformly throughout a cubical volume whose edges are 8.0 cm long. (a) What is the charge density in the cube? (b) What is the electric flux through a cube with 12 cm edges that is concentric with the charge distribution? 8 q 5 x10  C

2.

V  0.08m 

3

q  V 5x10 8 C  3  0.08 m

 9.77 x10  5

C m3

E E dA 

q 0

q All charge is enclosed by Gaussian surface





5 x10  8 C q in E   2 0    12 C x 8.85 10  2 Nm   Nm2 E 5649.72 C A charge of - 30 C is distributed uniformly throughout a spherical volume of radius 10 cm. Determine the electric field due to this charge at a distance of (a) 2 cm, (b) 5 cm, and (c) 20 cm from the center of the sphere

3.

q  30 x10 6 C 3 4 3 3 V    0.1m  4.19x10  m 3 r 0.1m q  V



  30 x10

 4.19 x10

 m 

6

  7.16 x10  3 a)

C

3

3

C m3

rR. r

++++

++++

r R a) qin  0  E 0 b)

rR

q EAg  0

qin   2  Rl 

  2 Rl  E  0  2 rl  R E rˆ  0r

5.

Charge is distributed throughout a spherical shell of inner radius r1 and outer radius r2 with a volume density given by  = or1/r, where o is constant. Determine the electric field due to this charge as a function of r, the distance from the center of the shell.

r1

r

r...