A) \[\frac{{{\theta }_{2}}-{{\theta }_{1}}}{{{\varepsilon }_{0}}}\]
B) \[\frac{{{\theta }_{1}}+{{\theta }_{2}}}{{{\varepsilon }_{0}}}\]
C) \[\frac{{{\theta }_{1}}-{{\theta }_{2}}}{{{\varepsilon }_{0}}}\]
D) \[{{\varepsilon }_{0}}\left( {{\theta }_{1}}+{{\theta }_{2}} \right)\]
Correct Answer: D
Solution :
Let \[-{{q}_{1}}\] be the charge, due to which flux\[\text{o }\!\!|\!\!\text{ }\] is entering the surface. \[\text{o}{{\text{ }\!\!|\!\!\text{ }}_{1}}=\frac{-q}{{{\varepsilon }_{0}}}\]or \[-{{q}_{1}}={{\varepsilon }_{0}}\text{o}{{|}_{1}}\] Let \[\text{+}{{\text{q}}_{\text{2}}}\]be the charge, due to which flux\[\text{o}{{\text{ }\!\!|\!\!\text{ }}_{2}},\] is entering the surface. \[\text{o}{{\text{ }\!\!|\!\!\text{ }}_{1}}=\frac{{{q}_{2}}}{{{\varepsilon }_{0}}}\]or \[-{{q}_{2}}={{\varepsilon }_{0}}\text{o}{{\text{ }\!\!|\!\!\text{ }}_{2}}\] Electric charge inside the surface \[={{q}_{2}}-{{q}_{1}}={{\varepsilon }_{0}}\text{o}{{\text{ }\!\!|\!\!\text{ }}_{2}}+{{\varepsilon }_{0}}\text{o}{{\text{ }\!\!|\!\!\text{ }}_{1}}\] \[={{\varepsilon }_{0}}(o{{|}_{2}}+o{{|}_{1}})\]
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