Question

A point charge q +3 HC is at the center of a sphere of radius 0.3 m. (a) Find the surface area of the sphere. (b) Find the magnitude of the electric field at points on the surface of the sphere. N/C (c) What is the flux of the electric field due to the point charge through the surface of the sphere? N m2/C (d) Would your answer to part (c) change if the point charge were moved so that it was inside the sphere but not at its center? O yes 0 no O not enough information to decide (e) What is the net flux through a cube of side 2 m that encloses the sphere? N. m2/C

Answer 1

"q = + 3 mu C = 3 times 10^-6 C radius of sphere r = 0.3 m (a) Surface area of sphere is S = 4 pi r^2 S = 4 times pi times (0.3)^2 m^2 = 1.131 m^2 (b) A point P on surface of sphere is at a distance r (radius) from charge at origin, so, from coulomb's law, Electric field at point p on surface of sphere is, E = 1/4 pi epsilon_0 q/r^2 N/C F = 9 times 10^9 times 3 times 10^-6/(0.3)^2 [Because 1/4 pi epsilon_0 = 9 times 10^9 Nm^2 c^-2 E = 3 times 10^5 N/C (c) Net flux through a surface (closed) is defined as 1/epsilon_0 times the net charge enclosed in that closed surface, which is ""Gauss law"". Therefore phi = q/epsilon_0 (1) [Also, Electric flux is number of electric field crossing a unit surface] So, phi is also defined as, phi contourintegral F vector middot ds vector"

I have simply used Gauss's law here which defines electric flux through a closed surface.

If you still have any confusion anywhere then please let me know in the comments section.

Thank you.