A charge q is enclosed by a spherical surface of radius r. if the radius if reduced to half, how would the electric flux through the surface change
Answers
Answered by
81
, inside the cube the
electric field is equal to
zero since the charge is
enclosed inside the cube
and the electric flux
remain the same because
the electric field inside the
cube affected the
expression and make it to
be equal to zero.
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electric field is equal to
zero since the charge is
enclosed inside the cube
and the electric flux
remain the same because
the electric field inside the
cube affected the
expression and make it to
be equal to zero.
Plz add me as a brainliest.
Answered by
29
Since we know that acc. To gauss theorem in electrostatics flux through a surface is equal to q/E° hence it does not depend on value of. R i.e. radius of Gaussian surface
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