Question

CI dz

In a cubical volume, 0.70 m on a side, the electric field is where E₀=0.125N/C and a=0.70m.

The cube has its sides parallel to the coordinate axes, see the figure. Determine the net charge within the cube.

Answer 1

side of the cubical volume = 0.70 m

a = 0.70 m

E₀= 0.125 N/C

$\underset{E}{\rightarrow}$ = 0.125 ( 1 + 0.70/0.70 ) i + 0.125 ( 0.70/0.70 )j

= 0.125 (2) i + 0.125 j

= 0.125( 2i + j)

for two opposite sides of the cube $\underset{E}{\rightarrow}$ = 0.125(2i+j)

therefore E total = 2* 0.125(2i+j)

$\underset{E}{\rightarrow}$_jk = 0.125(0.70/0.70)j

= 0.125 j

  $\underset{E}{\rightarrow}$_jk takes place on two sides so it is equal to = 2*0.125 j

  $\underset{E}{\rightarrow}$_ki = 0.125 ( 1+ 0.70/0.70)i

  $\underset{E}{\rightarrow}$_ki= 0.125(2i)

this takes place on two sides therefore    $\underset{E}{\rightarrow}$ _ki = 2*0.125 *2i

so the net field=   2*0.125 *(2i +j) +   2*0.125 *(j) + 2*0.125 *(2i)

= 0.250 (2i + j+ j + 2i)

= 0.250 ( 4i + 2j )

=0.500(2i+ j)