Question

Find the electric flux through each face of a hollow cube of side 10 cm, if a charge of 8.85 µC is placed at the centre.

Answer 1

Hey There!!



Here, we are given that a charge of 8.85 $\mu C$ is placed at the center of a cube.


Let us first find the total flux with the help of Gauss' Law. 


Gauss Law states that:

The total flux through any closed surface is equal to the charge enclosed by the closed surface divided by the absolute permittivity.


In mathematical form, we have:

$\phi = \frac{q_{in}}{\varepsilon_{\circ}}$


Let us find the total flux through the cube.


$\phi = \frac{q_{in}}{\varepsilon_{\circ}} \\ \\ \\ \implies \phi = \frac{8.85 \times 10^{-6}}{8.85 \times 10^{-12}} \\ \\ \\ \implies \phi = 10^6 \, \, N \, m^2 \, C^{-1}$


Now, we know the total flux through the cube. Since the charge is placed at the center, the flux through each face of cube must be same.

Now, a cube has six faces. And flux thorugh each face is same. 


So, Flux through each face must be one-sixth of the total flux.


$\implies \text{Flux through each face }=\frac{\phi}{6} \\ \\ \\ \implies \text{Flux through each face }= \frac{10^6}{6} \\ \\ \\ \implies \boxed{\text{Flux through each face } \approx 1.67 \times 10^5 \, \, N \, m^2 \, C^{-1}}$


Hope it helps
Purva
Brainly Community

Answer 2

The electric flux through each face of a hollow cube of side 10 cm:

Charge =q=8.85μC=8.85 x 10⁻⁶C is placed at the centre

Flux linked with each face of cube =Φ/6

where Φ=q/ε₀

 Flux linked with each face of cube=q/6xε₀

=1/6[8.85 x 10⁻⁶/8.85 x10⁻¹²]

=10⁶/6

=0.16 x10⁶ Nm²C⁻¹

=1.6 x 10⁵ Nm²C⁻¹


∴Flux linked with each face of cube is 1.6 x 10⁵ Nm²C⁻¹