Question

3. a) Find the net electric flux through this cube. b) Explain why we can't use Gauss's Law to find the electric field here. -8.00 nC +3.00 nC c) Is there a shape to which Gauss's Law could be applied in order to solve for the electric field of this double point charge situation?

Answer 1

solution S WO -8.oonc s Using Gauss's theorem The net electric Flux through this cube is ginen by 0 = & Eid A = a inclosed or 6- 3x10 Ic 8.85 81012 CM 0 = 0.33 9 x103 0 3.39 x 10 only It we use Gauss's how to compute electric field here, we would get field, due to charge 22=+3nc while due to charge a, will be letht. Because - law is applied for only due to closed charges not outside of the closed surface. That's why we can't use Gauss's how to find electric field here. By Yes, it is a cylinder If we use a cylider both charges will be lying inside the closed surface & we can easily use auss's law to solue for electric field of this double pt. charge situation. Any +3nc

Note - In above solection we are not taking into account flux due to charge ai. For this there are two reasons 0 Gauss's law says that only closed charges will be considerable in computing flux. If we use a, por comple Computing flux. It will produce two flux, firstly, Ilux due electric field lines entering the surface & (say) & secondly, flux due to field lines leaving the surface is – 6, Then net flux due to charge q, is 0= 6 +60)
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