A) \[{{\varepsilon }_{0}}({{\phi }_{1}}+{{\phi }_{2}})\]
B) \[\frac{{{\phi }_{2}}+{{\phi }_{1}}}{{{\varepsilon }_{0}}}\]
C) \[\frac{{{\phi }_{1}}-{{\phi }_{2}}}{{{\varepsilon }_{0}}}\]
D) \[\frac{{{\phi }_{1}}+{{\phi }_{2}}}{{{\varepsilon }_{0}}}\]
Correct Answer: A
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 be the charge, due to which the flux \[\phi \], is entering the surface, \[{{\phi }_{1}}=-\frac{{{q}_{1}}}{{{\varepsilon }_{0}}}\] \[\Rightarrow \] \[-{{q}_{1}}={{\varepsilon }_{0}}{{\phi }_{2}}\] Let + \[{{q}_{2}}\]be the charge, due to which the flux \[{{\phi }_{2}}\] is leaving the surface \[\therefore \] \[{{\phi }_{2}}=\frac{{{q}_{2}}}{{{\varepsilon }_{0}}}\] \[\Rightarrow \] \[q={{\varepsilon }_{0}}{{\phi }_{2}}\] So, charge inside the surface \[={{q}_{2}}-{{q}_{1}}\] \[={{\varepsilon }_{0}}{{\phi }_{2}}+{{\varepsilon }_{0}}{{\phi }_{1}}={{\varepsilon }_{0}}({{\phi }_{1}}+{{\phi }_{2}})\]
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