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Question: The unit of permittivity is: \(\begin{aligned} & \text{A}\text{. }{{\text{C}}^{2}}{{\text{N}}^...

The unit of permittivity is:
A. C2N1m2 B. Nm2C2 C. Hm1 D. NC2m2 \begin{aligned} & \text{A}\text{. }{{\text{C}}^{2}}{{\text{N}}^{-1}}{{\text{m}}^{-2}} \\\ & \text{B}\text{. N}{{\text{m}}^{2}}{{C}^{-2}} \\\ & \text{C}\text{. H}{{\text{m}}^{-1}} \\\ & \text{D}\text{. N}{{\text{C}}^{2}}{{\text{m}}^{-2}} \\\ \end{aligned}

Explanation

Solution

Hint: Unit of permittivity can be deduced from coulomb's law. Use the formula of coulomb's law of force of attraction between two charges. You will find permittivity in a constant of proportionality. If charges are measured in coulomb, r measured in meter and forces are measure in Newton then ε0=8.85×1012C2Nm2{{\varepsilon }_{0}}=8.85\times {{10}^{-12}}\dfrac{{{C}^{2}}}{N{{m}^{2}}}.

Complete step by step solution:
We know that coulomb's law measured electrical attraction and repulsions quantity.
So coulomb's law states that the electrostatic force of attraction or repulsion between two point electric charges at rest is directly proportional to the product of the magnitude of two charges and inversely proportional to the square of the distance between them. The force acts along the line joining the charges.
Let q1 and q2 are two charges separated by distance r and force of attraction between them is F.
Mathematically,
Fαq1q2r2F\alpha \dfrac{q1q2}{{{r}^{2}}}
Where, K is constant of proportionality.
If charges are kept in vacuum the, K is expressed as
K=14πε0K=\dfrac{1}{4\pi {{\varepsilon }_{0}}}
Where ε0{{\varepsilon }_{0}}is the permittivity of free space.
Then Coulomb force is given by,
F=14πε0q1q2r2 ε0=q1q2Fr2 \begin{aligned} & F=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{q1q2}{{{r}^{2}}} \\\ & {{\varepsilon }_{0}}=\dfrac{q1q2}{F{{r}^{2}}} \\\ \end{aligned}
Unit of force F= Newton (N)
Unit of charge q1 and q2= Coulomb(C)
Unit of r= meter (m)
Then,unit of ε0=C2Nm2=C2N1m2\text{unit of }{{\varepsilon }_{0}}=\dfrac{{{C}^{2}}}{N{{m}^{2}}}={{C}^{2}}{{N}^{-1}}{{m}^{-2}}
So the unit of permittivity is given by C2N1m2{{C}^{2}}{{N}^{-1}}{{m}^{-2}}.

Answer is (A)

Note: Unit of Permittivity is also farad/m. K also can be returned as K=εε0K=\dfrac{\varepsilon }{{{\varepsilon }_{0}}}, where ε\varepsilon is the permittivity of the medium. Permittivity of medium or dielectric constant of medium measures the ability of the medium to store energy in the electric field in it. It also depends upon the temperature and other physical conditions.
The medium modifies the field, force, potential energy etc. Every medium shows different modification and thus has different values of permittivity.
Coulomb scientists in 1785, first measured electrical attraction with repulsion and we know that if the charges of two bodies are like that is both positive or negative, the body repels each other whereas unlike attracts each other.