What is the net flux of the uniform electric field of question 1.15 through a cube of side 20 cm oriented so that its faces are parallel to the coordinate planes?

Given,

straight E with rightwards arrow on top space equals space 3 space cross times space 10 cubed space straight i with hat on top space NC to the power of negative 1 end exponent

The area of each face out of the six faces of the cube = 20 x 20 = 400 cm² = 4 x 10^–2 m². The area vector of four faces of cube ABGH, OBGF, OCDF and ACDH is along +Y axis, –Z axis, –Y axis and +Z axis. The direction of E (+X axis).

stack increment straight S with rightwards arrow on top

is perpendicular to each other so flux through there surfaces are zero.
Given,The area of each face out of the six faces of the cube = 20 x

Hence, the net electric flux through the cube

straight ϕ space equals space straight E with rightwards arrow on top. space increment stack straight S subscript 1 with rightwards arrow on top space plus space straight E with rightwards arrow on top. space increment space stack straight S subscript 2 with rightwards arrow on top

Now,

open vertical bar increment stack straight S subscript 1 with rightwards arrow on top close vertical bar space equals space open vertical bar increment stack straight S subscript 2 with rightwards arrow on top close vertical bar space equals space 4 cross times 10 to the power of negative 2 end exponent straight m squared

and the angle between

straight E with rightwards arrow on top space and space increment stack straight S subscript 1 with rightwards arrow on top

0 degree comma

whereas the angle between

straight E with rightwards arrow on top space and space increment stack straight S subscript 2 with rightwards arrow on top space is space 180 degree

Thus,

straight ϕ space equals space straight E space increment straight S subscript 1 space cos space 0 degree space space plus straight E space increment straight S subscript 2 space cos space 180 degree space space space equals space straight E cross times 4 cross times 10 to the power of negative 2 end exponent cross times 1 plus straight E cross times 4 cross times 10 to the power of negative 2 end exponent cross times left parenthesis negative 1 right parenthesis space space space space equals space 0

Now, it is established that if some electric flux enters the cube the same amount of flux leaves through the other face, so that the net flux is zero.

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