Question

An electric flux of -6 x 10^3 nm^2c^-1 passes through a spherical gaussian surface of radius 10 cm due to a point charge placed at the centre.(i)what is the charge enclosed by this gaussian surface?(ii) if the radius of the gaussian surface is doubled,how much flux would pass through the surface?

Answer 1

$$\begin{gathered} i)AccordingtoGauss\text{'}slawofelectricflux\left(\Phi \right)={\oint }_s\overrightarrow{E}.\overrightarrow{S}=\frac{q}{{\epsilon }_0},where\overrightarrow{E}iselectricfield,\overrightarrow{S}thesurfacearea,qchargeeclosedbyGaussiansurfaceand \\ {\epsilon }_{0}representpermittivityoffreespace. \\ Wearegivenelectricflux\left(\Phi \right)={\oint }_s\overrightarrow{E}.\overrightarrow{S}=-6\times {10}^{3}N{m}^2{C}^{-1}.Therefore, \\ \frac{q}{{\epsilon }_0}=-6\times {10}^{3}orq=-6\times {10}^3\times {\epsilon }_{0}orq=-6\times {10}^3\times 8.85\times {10}^{-12}C. \\ Orq=-53.10\times {10}^{-9}CORq=-5.31\times {10}^{-8}C. \\ ii)Thefluxpas\sin gthroughthesphericalGaussiansurfaceofradius20cmisthesame-6\times {10}^{3}N{m}^2{C}^{-1}. \end{gathered}$$