Electrostatics Question 733
Question: The inward and outward electric flux for a closed surface in units of $ N-m^{2}/C $ are respectively $ 8\times 10^{3} $ and $ 4\times 10^{3}. $ Then the total charge inside the surface is [where $ {\varepsilon _{0}} $ = permittivity constant]
Options:
A) $ 4\times 10^{3}C $
B) $ 3.14Nm^{2}/C $
C) $ \frac{(-4\times 10^{3})}{\varepsilon }C $
D) $ -4\times 10^{3}{\varepsilon _{0}}C $
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Answer:
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{.} $
