A charge Q is enclosed by a Gaussian spherical surface of radius R. If the radius is doubled, then the outward electric flux will

CONCEPT:

• Gauss's Law: Gauss's law for the electric field describes the static electric field generated by a distribution of electric charges. 
  • It states that the electric flux through any closed surface is proportional to the total electric charge enclosed by this surface. 

\(\phi_E = \frac{Q}{\epsilon_o}\)

Where ϕ_E = electric flux through a closed surface S enclosing any volume V, Q = total charge enclosed with V, and ϵ_o = electric constant

• Electric flux is the flow of the electric field through a given area.

• Electric flux is proportional to the number of electric field lines going through a virtual surface.

\({\phi}_E= E\cdot S = EScos\theta\)

Where E = electric field, S = are of the surface, E = magnitude, θ = angle between the electric field lines and the normal (perpendicular) to S, and ϕ_E = flux the electric field through a closed cylindrical surface.

EXPLANATION:

• According to Gauss's law

 \(ϕ_E = \frac{Q_(enclosed )}{ϵ_o}\)

• If the radius of the Gaussian surface is doubled, the outward electric flux will remain the same.
• This is because electric flux depends only on the charge enclosed by the surface.

​option 4 is correct.