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{.} $



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