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When the electric flux associated with closed surface becomes positive, zero or negative ? |
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Answer» Solution :If electric flux ASSOCIATED with area s due to electric field E is `vecE`, then `phi = vecE. vecS` `therefore vecphi = ES cos theta`……(1) Where `theta` is angle between `vecE` and `vecS`. (i) If `vecS bot vecE` MEANS surface is parallel to `vecE` then, `theta = 90^(@)` `therefore` from equation (1) `phi = ES cos 0^(@)=0` `therefore` Flux associated with surface is zero. (ii) If `theta lt 90^(@)`, then `cos theta GT 0` (negative) hence flux is POSITIVE. (iii) If `theta gt 90^(@)`, then `cos theta lt 0`. (negative) hence flux is negative. All three are shown in FIGURE (a), (b) and (c) respectively. 
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