Electric flux φ through a closed surface enclosing charge Q is Q ε0 . What will be the electric flux through the closed surface if its size is doubled? Please answer only if you know.
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It will remain Q/epsilonnot because by Gauss' Law, Electric Flux only depends on the charge enclosed by the Gaussian surface and does not depend on the radius . Hence , Electric Flux will not change.
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- Electric flux is the measure of electric field through given surface area
∅ = E.S = ES cosθ
- Electric flux depends on the magnitudes of the electric field and the area, as well as the relative orientation of the area with respect to the direction of the electric field.
- As per the Guess theorem in electrostatics, electric flux does not depend on the shape or size of the surface. The electric flux depends only on the charge enclosed by the surface.
Therefore the electric flux will be same
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