Q 1.20(a): A point charge causes an electric flux of -1.0\times 10^{3}\frac{Nm^{2}}{C}to pass through a spherical Gaussian surface of 10.0 cm radius centred on the charge. If the radius of the Gaussian surface were doubled, how much flux would pass through the surface?

Answers (1)

Given,

\phi = -1.0\times 10^{3}\frac{Nm^{2}}{C}

Using Gauss’s law, we know that the flux of electric field through any closed surface S is 1/\epsilon _{0}  times the total charge enclosed by S.

i.e.   \phi = q/\epsilon _{0}

where, q = net charge enclosed  and  \epsilon _{0} = permittivity of free space (constant)

Therefore, Flux does not depend on the radius of the sphere but only on the net charge enclosed. Hence, the flux remains the same although radius is doubled.

\phi' = -10^{3}\frac{Nm^{2}}{C}

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