Question

A cube of side 3.56 cm has a charge of 9.11 μ C placed at its center. Calculate the electric flux through one side of the cube

Answer 1

Flux=q/e0
e0=8.85*10^-12
!!! Cube has 6 sides, so flux through one side is equal to total flux/6
Ans=9.11*10^(-6)/((8.85*10(-12))*6)=(approximately)1.72*10^5Nm^2/C

Answer 2

`\phi_(surface) = 1.72 * 10^5 N/Cm^2`

Step-by-step explanation:

As we know by the theory of flux that if charge "q" is enclosed in a closed surface then total flux through the surface is given as

`\phi = (q)/(\epsilon_0)`

so here total flux passing through the cube is given as

`\phi_(total) = (9.11\muC)/(8.85 * 10^(-12))`

now we will have

`\phi_(total) = 1.03* 10^6`

now as we know that charge is placed at the center of the cube

so the total flux is uniformly passing through all faces of the cube

so flux passing one side of the cube is given as

`\phi_(surface) = (1.03* 10^6)/(6)`

`\phi_(surface) = 1.72 * 10^5 N/Cm^2`