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