I have a doubt sir doesn't total flux should account for 2 surfaces therefore the closed integral should be once e.dacos180 and e.dacos0 as at one side arew vector is perpendicular to e and one side it is parallel to e
I'm not sure what you're getting at here, but to summarize the calculation of flux: the integrals over the upper surface and over the lower surface will both yield the same positive flux because E is parallel to the normal vector (normal vector will point perpendicular to the incremental area and away from the inside of the Gaussian surface), the field magnitude is the same, and the area is the same. So I used a factor of 2 to account for the total flux through the upper and lower surfaces. The flux through the sides is zero since the electric field does not penetrate through the sides (it is parallel to the surface/perpendicular to the normal vector there).
respected sir , which type of software are you using sir for writing and screen recording ??
Powerpoint for all graphics with a plugin called MathType for equations. OBS for screen capture.
excellent sir
thanks! - Zak
I have a doubt sir doesn't total flux should account for 2 surfaces therefore the closed integral should be once e.dacos180 and e.dacos0 as at one side arew vector is perpendicular to e and one side it is parallel to e
I'm not sure what you're getting at here, but to summarize the calculation of flux: the integrals over the upper surface and over the lower surface will both yield the same positive flux because E is parallel to the normal vector (normal vector will point perpendicular to the incremental area and away from the inside of the Gaussian surface), the field magnitude is the same, and the area is the same. So I used a factor of 2 to account for the total flux through the upper and lower surfaces. The flux through the sides is zero since the electric field does not penetrate through the sides (it is parallel to the surface/perpendicular to the normal vector there).