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, \[{{\phi }_{total}}\,=\frac{\theta }{{{\varepsilon }_{0}}}\] Let electric flux linked with surfaces A, B and C are \[{{\phi }_{A}},\] \[{{\phi }_{B}}\] and \[{{\phi }_{C}}\] respectively That is \[{{\phi }_{total}}\,={{\phi }_{A}}+{{\phi }_{B}}+{{\phi }_{C}}\] \[{{\phi }_{C}}={{\phi }_{A}}\] \[2{{\phi }_{A}}={{\phi }_{_{total}}}\,=\frac{q}{{{\varepsilon }_{0}}}\] Or \[{{\phi }_{A}}=\frac{1}{2}\,\left( \frac{q}{{{\varepsilon }_{0}}}\,-{{\phi }_{B}} \right)\] But \[{{\phi }_{B}}=\phi \] (given) \[{{\phi }_{A}}=\frac{1}{2}\left( \frac{q}{{{\varepsilon }_{0}}}-\phi \right)\]
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