What is the net flux of the uniform electric field of Exercise 1.15 through a cube of side 20 cm oriented so that its faces are parallelto the coordinate planes?
Given: The side of the cube is 20 cm and the electric field is 3 × 10 3 i ^ NC − 1 .
The area of each side of the cube is given as,
A = s 2
Where, s is the length of the side of the cube.
By substituting the given values in the above equation, we get
A = ( 0.2 ) 2 = 0.04 m 2
Consider the diagram below.
The net flux through the cube is given as,
ϕ = ∑ | E | A cos θ
Since, the area vectors of the four faces BCFE , OADG , DEFG and OABC is perpendicular to the electric field vector, so the flux through these surfaces will be zero.
By substituting the values in the above expression, we get
ϕ = E A 1 cos 0 ° + E A 2 cos 180 ° = ( 3 × 10 3 ) ( 0.04 ) ( 1 ) + ( 3 × 10 3 ) ( 0.04 ) ( − 1 ) = 0
Thus, the net flux through the cube is zero.
Given: The side of the cube is
The area of each side of the cube is given as,
Where,
By substituting the given values in the above equation, we get
Consider the diagram below.
The net flux through the cube is given as,
Since, the area vectors of the four faces
By substituting the values in the above expression, we get
Thus, the net flux through the cube is zero.
