A) \[4\times {{10}^{3}}C\]
B) \[3.14\,\,N{{m}^{2}}/C\]
C) \[\frac{(-\,4\times {{10}^{3}})}{\varepsilon }C\]
D) \[-\,4\times {{10}^{3}}{{\varepsilon }_{0}}C\]
Correct Answer: D
Solution :
[d] By Gauss's law \[\phi =\frac{1}{{{\varepsilon }_{0}}}({{Q}_{emclosed}})\] \[\Rightarrow {{Q}_{emclosed}}=\phi {{\varepsilon }_{0}}=(-8\times {{10}^{3}}+4\times {{10}^{3}})\,{{\varepsilon }_{0}}\] \[=-4\times {{10}^{3}}{{\varepsilon }_{0}}\text{ coulomb}\text{.}\]
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