Question

The value of electrical permittivity of free space is
A) $9\times {10}^{9}N{C}^2/m^2$
B) $8.85\times {10}^{-12}N{m}^2/C^{2}s$
C) $8.85\times {10}^{-12}{C}^2/N-m^2$
D) $9\times {10}^9{C}^2/N-m^2$

Answer 1

Hint: In an ammeter, a galvanometer and a shunt resistance are connected in parallel. The current that will give full-scale deflection in the absence of the shunt is nearly equal to the current through the galvanometer when the shunt is connected.

Complete step by step answer:
The physical constant ${\epsilon }_0$ , the vacuum permittivity, is the permittivity of free space or electric constant or the distributed capacitance of the vacuum. It is a physical constant, which is the value of the absolute dielectric permittivity of classical vacuum. The vacuum permittivity is also related to the speed of light in the vacuum.
It represents the capacity of a medium to allow electric fields to sustain in the medium. It is also related to the energy stored within an electric field.
The value of the ${\epsilon }_0$ is
 ${\epsilon }_0=8.85\times {10}^{-12}{C}^2/N-m^2$
This is the permittivity of free space.
Hence the correct choice is option (C).

Note:
The units of the permittivity constant can be determined from coulomb’s law which relates the force between two charged objects. It says
 where $F$ is the force between two charged objects $q_1,q_2$ at a distance $r$ in a vacuum.
In the above equation, the dimensions of force are in $(N)$ Newton’s, the product of the two charges will be $C^2$ and the units of distance squared will be $m$ (metre). Hence the units of permittivity will be $C^2/N-m^2$ .
Option (D) can be confused with the answer as well. But the value in option (D) corresponds to which is constant, used in Coulomb’s law so we should not confuse the two.