A) \[\frac{{{\phi }_{2}}-{{\phi }_{1}}}{{{\varepsilon }_{0}}}\]
B) \[\frac{{{\phi }_{1}}+{{\phi }_{2}}}{{{\varepsilon }_{0}}}\]
C) \[\frac{{{\phi }_{1}}-{{\phi }_{2}}}{{{\varepsilon }_{0}}}\]
D) \[{{\varepsilon }_{0}}\,({{\phi }_{1}}+{{\phi }_{2}})\]
Correct Answer: D
Solution :
According to this law, the net electric flux through any dosed surface is equal to the net charge inside the surface divided by \[{{\varepsilon }_{0}}.\] Therefore, \[\phi \text{=}\frac{q}{{{\varepsilon }_{0}}}\] Let \[-{{q}_{1}}\]be the charge, due to which flux \[{{\phi }_{1}}\] is entering the surface \[\therefore \] \[\phi \text{=}\frac{-{{q}_{1}}}{{{\varepsilon }_{0}}}\] or \[-{{q}_{1}}={{\varepsilon }_{0}}{{\phi }_{1}}\] Let \[+\,{{q}_{2}}\]be the charge, due to which flux \[{{\phi }_{2}}\] is leaving, the surface \[\therefore \] \[{{\phi }_{2}}=\frac{{{q}_{2}}}{{{\varepsilon }_{0}}}\] or \[{{q}_{2}}={{\varepsilon }_{0}}{{\phi }_{2}}\] So, electric charge inside the surface \[={{q}_{2}}-{{q}_{1}}\] \[={{\varepsilon }_{0}}{{\phi }_{2}}+{{\varepsilon }_{0}}{{\phi }_{1}}\] \[={{\varepsilon }_{0}}({{\phi }_{2}}+{{\phi }_{1}})\]
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