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

Answer:

$\phi_{surface} = 1.72 \times 10^5 N/Cm^2$

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

$\phi = \frac{q}{\epsilon_0}$

so here total flux passing through the cube is given as

$\phi_{total} = \frac{9.11\muC}{8.85 \times 10^{-12}}$

now we will have

$\phi_{total} = 1.03\times 10^6$

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

$\phi_{surface} = \frac{1.03\times 10^6}{6}$

$\phi_{surface} = 1.72 \times 10^5 N/Cm^2$

Answer 2

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