The electric flux from a cube of side ‘a’ is ‘Φ’. What will be its value if the side of the cube is made ‘2a’ and the charge enclosed is made half?

The correct answer is option 1) i.e. Φ/2

CONCEPT:

Gauss's Law for electric field: It states that the total electric flux emerging out of a closed surface is directly proportional to the charge enclosed by this closed surface. It is expressed as:

\(ϕ = \frac{q}{ϵ_0}\)

Where ϕ is the electric flux, q is the charge enclosed in the closed surface and ϵ₀ is the electric constant.

EXPLANATION:

Given that:

Consider a charge 'q' placed inside a cube of side 'a'.

The electric flux according to Gauss's law, 

\(\Rightarrow ϕ = \frac{q}{ϵ_0}\)

If the charge enclosed is halved, then  

\(\Rightarrow q' =\frac{q}{2}\)

Therefore, the new electric flux associated with this, 

\(\Rightarrow ϕ' =\frac{q'}{\epsilon_0}= \frac{\frac{q}{2}}{ϵ_0} = \frac{1}{2}\times \frac{q}{\epsilon_0} \)

\(\Rightarrow \phi '= \frac{1}{2} \times \phi = \frac{\phi}{2}\)

The electric flux emerging from a closed surface is independent of the shape or dimensions of the closed surface.