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(i) In figure (a) calulate the electric flux through the closed areas A_(1) and A_(2) (ii) In figure (b) calculate the electric flux through the cube |
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Answer» Solution :(i) In figure (a) the area `A_(1)` encloses the charge Q. So ELECTRIC flux through this closed surface `A_(1)` is `(Q)/(epsilon_(0))` . But the closed surface `A_(2)` contains no charges inside so electric flux through `A_(2)` is zero . (ii) In figure (b) the net charge inside the cube is 3p and the TOTAL electric flux in the cube is therefore `Phi_(E)=(3q)/(epsilon_(0))` NOTE that the charge `-10q` lies outside the cube and it will not contribute the total flux through the surface of the cube. |
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