Question

a point charge of magnitude 8.85x10^-12C is pkaced at the centre of a cube. electric flux passing through one face​

Answer 1

Answer:

1/6

Explanation:

By Gauss's law,

• electric flux enclosed by the total surface(6 faces) of the cube=q/£•
• therefore, electric flux through one face

=q/6£•

=8.85×10^-12/(6×8.75×10^-12)

=1/6

(note: value of £•=8.85×10^-12)

hope this helps..!

Answer 2

Given : a point charge of magnitude 8.85x10^-12C is placed at the centre of a cube.

To find : The electric flux passing through one face.

solution : From Gauss law, electric flux through the Gaussian surface is the ratio of charge enclosed the surface to the permittivity of medium. i.e., Φ = Q/ε

here a point charge is placed at the centre of a cube ( Gaussian surface).

so, Total electric flux passing through the cube, Φ = q/ε

here, q = 8.85 × 10¯¹² C, ε = 8.85 × 10¯¹² C²/Nm²

so, Φ = (8.85 × 10¯¹² C)/(8.85 × 10¯¹² C²/Nm²)

= 1 Nm²/C

as it is being shared by six faces of cube.

so, flux passing through each face of cube = Φ/6

= 1/6 Nm²/C

Therefore The electric flux passing through one face is 1/6 Nm²/C.