# The electric flux over a sphere of radius 1m is  A . If radius of the sphere were doubled without changing the charge enclosed electric flux would become ?

419 Points
11 years ago

Dear javed

For a volume V with surface S, Gauss's law states that

$\Phi_{E,S} = \frac{Q}{\varepsilon_0}$

where ΦE,S is the electric flux through S, Q is total charge inside V, and ε0 is the electric constant

since Q=constant flux will remain same equal to A

All the best.

AKASH GOYAL

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509 Points
11 years ago

flux is independent of area of sphere & it depends only on charge enclosed by the surface so

the sphere whose radius is double will have same flux A...