A) \[\frac{1}{2}\left( \frac{q}{{{\varepsilon }_{0}}}-\phi \right)\]
B) \[\frac{q}{2{{\varepsilon }_{0}}}\]
C) \[\frac{\phi }{3}\]
D) \[\frac{q}{{{\varepsilon }_{0}}}-\phi \]
Correct Answer: A
Solution :
According to Gauss's law the net electric flux through any closed surface is equal to the net charge inside the surface divided by\[{{\varepsilon }_{0}}.\]That is \[{{\phi }_{total}}=\frac{q}{{{\varepsilon }_{0}}}\] Let electric flux linked with surfaces A, B and C are\[{{\phi }_{A}},{{\phi }_{B}}\]and \[{{\phi }_{C}}\]respectively. That is \[{{\phi }_{total}}={{\phi }_{A}}+{{\phi }_{B}}+{{\phi }_{C}}\] Since \[{{\phi }_{C}}={{\phi }_{A}}\] \[2{{\phi }_{A}}+{{\phi }_{B}}={{\phi }_{total}}=\frac{q}{{{\varepsilon }_{0}}}\] or \[{{\phi }_{A}}=\frac{1}{2}\left( \frac{q}{{{\varepsilon }_{0}}}-{{\phi }_{B}} \right)\] But \[{{\phi }_{B}}=\phi \](given) Hence, \[{{\phi }_{A}}=\frac{1}{2}\left( \frac{q}{{{\varepsilon }_{0}}}-\phi \right)\]
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