A) \[\frac{\pi q}{6(4\pi {{\varepsilon }_{0}})}\]
B) \[\frac{q}{6(4\pi {{\varepsilon }_{0}})}\]
C) \[\frac{2\pi q}{6(4\pi {{\varepsilon }_{0}})}\]
D) \[\frac{4\pi q}{6(4\pi {{\varepsilon }_{0}})}\]
Correct Answer: D
Solution :
| [d] Key Idea: Gauss's law states that, the net electric flux through any closed surface is equal to the net charge inside the closed surface divided \[{{\varepsilon }_{0}}\]. |
| Thus, from Gauss's law |
| \[\phi =\frac{q}{{{\varepsilon }_{0}}}\] |
| This is the net flux coming out from cube. |
| Since, a cube has \[\phi \] sides so electric flux through any face is |
| \[\phi '=\frac{\phi }{6}=\frac{q}{6{{\varepsilon }_{0}}}\] |
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