A) \[8.89\times {{10}^{-11}}C\]
B) \[4.47\times {{10}^{-11}}C\]
C) \[8.89\times {{10}^{-9}}C\]
D) \[4.47\times {{10}^{-9}}C\]
Correct Answer: A
Solution :
[a] : The electric field at the surface of the sphere is Aa and being radial it is along the outward normal. The flux of the electric field is, therefore, \[\Phi =\oint{EdS\cos {{0}^{o}}}=Aa(4\pi {{a}^{2}})\]. From Gauss?s law, the charge contained in the sphere is \[{{Q}_{inside}}={{\varepsilon }_{0}}\Phi =4\pi {{\varepsilon }_{0}}A{{a}^{3}}\] \[=\frac{1}{9\times {{10}^{9}}}\times 100V\,{{m}^{-2}}\times {{(0.20)}^{3}}\] \[=8.89\times {{10}^{-11}}C\]
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