Question

electric field components are Ex = 100 x^1/2 Ey= Ez=0 . Calculate net flux through the cube placed in electric field at shown position.​

Answer 1

$E_{X} = 100 {x}^{ \frac{1}{2} }$

$E_{Y} = 0 \: and \: E_{Z} = 0$

Now, let's assume a cube (of edge L) as shown in attached diagram:

• Flux will be only through the shaded surfaces.

Net flux will be :

$\Delta \phi = \phi_{EFGH} - \Delta \phi_{ABCD}$

$\implies \Delta \phi = 100 {L}^{ \frac{1}{2} } - 100( {0}^{ \frac{1}{2} } )$

$\implies \Delta \phi = 100 {L}^{ \frac{1}{2} } - 0$

$\implies \Delta \phi = 100 {L}^{ \frac{1}{2} }$

So, net flux is 100L^(1/2).