If the charge remains the same but the radius of the sphere is doubled, the electric flux coming out of it will be

Explanation:

We shall apply Gauss's theorem for electric flux to solve the problem . According to this theorem , total electric flux coming out of a charge q can be given by the following relation .

∫ E ds = q / ε

Here q is assumed to be enclosed in a closed surface , E is electric intensity on the surface so

∫ E ds represents total electric flux passing through the closed surface due to charge q enclosed in the surface .

This also represents total flux coming out of the charge q on all sides .

This is equal to q / ε where ε is a constant called permittivity  which depends upon the medium enclosing the charge . For air , its value is 8.85 x 10⁻¹² .

If charge remains the same but radius of the sphere enclosing the charge is doubled , the flux coming out of charge will remain the same .

It is so because flux coming out of charge q is q / ε . It does not depend upon surface area enclosing the charge . It depends upon two factors

1 ) charge q and

2 ) the permittivity of medium  ε  around .

see earlier answer

it would be c. predictability