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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 ?


5 years ago


Answers : (2)


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.




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5 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...

5 years ago

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