The electric flux
(Φ) through a closed surface, such as the surface of a cube, is directly proportional to the charge enclosed
(Q) by that surface. This relationship is dictated by Gauss's Law, which is represented by the equation:
Φ=‌where:
Φ is the electric flux,
Q is the charge enclosed by the surface,
ε0 is the permittivity of free space (a constant).
In the given problem,
The initial flux from a cube of side
1m is
Φ.
When the side of the cube is increased to
3m, and the charge enclosed is reduced to a third of its original value
(Q′=‌.) the relationship above can be used to find the new flux
(Φ′).
Using the relation for electric flux and substituting the new charge value, we get:
Φ′=‌=‌=‌⋅‌Since the original flux
Φ=‌, substituting this back into the equation for
Φ′ yields:
Φ′=‌⋅ΦThus, the electric flux from the larger cube with side
3m and one third the original enclosed charge is
‌. Therefore, the correct answer is:
Option A:
‌