Question

The unit of permittivity of free space $ \left( {{\varepsilon _0}} \right) $ is:
(A) $ C{N^{ - 1}}{m^{ - 1}} $
(B) $ N{m^{ - 2}}{C^{ - 2}} $
(C) $ {C^2}{N^{ - 2}}{m^{ - 2}} $
(D) $ {C^2}{N^{ - 1}}{m^{ - 2}} $

Answer 1

Hint Electric field is directly proportional to charge and inversely proportional to distance to the square of the distance from the charge. It has a constant of proportionality that is inversely related to the permittivity.

Formula used: $ E = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{Q}{{{r^2}}} $ where $ Q $ is charge measured in coulombs, $ r $ is distance measured in meters, and $ E $ is electric field measured in Newton per coulomb.

Complete step by step answer
In general, the permittivity of free space is a measure of the resistance of vacuum to the effect of electric field. Permittivity also exists for other materials or medium electric field can be felt or propagated. In summary, the higher the permittivity of the medium the lower the electric field effect for the same charge and distance considered from charge.
To find the unit of permittivity we can use the formula for the electric field and make $ {\varepsilon _0} $ subject of the formula. Then work with the variables’ unit to reveal the unit of $ {\varepsilon _0} $ .
The expression for electric field is given as
 $\Rightarrow E = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{Q}{{{r^2}}} $ where $ Q $ is charge measured in coulombs, $ r $ is distance measured in meters, and $ E $ is electric field measured in Newton per Coulomb.
Making $ {\varepsilon _0} $ subject of the formula, by multiplying both sides by $ {\varepsilon _0} $ and dividing both sides by $ E $ , we have that
$\Rightarrow {\varepsilon _0} = \dfrac{1}{{4\pi E}}\dfrac{Q}{{{r^2}}} $ . Therefore, the unit of $\Rightarrow {\varepsilon _0} $ will be given as the unit of the right hand side of the equation.
Hence, replacing each variable with its unit, we have that,
$\Rightarrow {\varepsilon _{0U}} = \dfrac{C}{{\dfrac{N}{C} \times {m^2}}} = C \div \left( {\dfrac{N}{C} \times {m^2}} \right) $ (since $ 4\pi $ is a dimensionless constant), where $ {\varepsilon _{0U}} $ is the unit of $ {\varepsilon _0} $ .
Evaluating the right hand side (by converting the division to multiplication) we have
$\Rightarrow {\varepsilon _{0U}} = C \times \left( {\dfrac{C}{{N{m^2}}}} \right) = {C^2}{N^{ - 1}}{m^{ - 2}} $
 $ \therefore {\varepsilon _{0U}} = {C^2}{N^{ - 1}}{m^{ - 2}} $
Hence, the correct option is D.

Note
Alternatively, to find the unit of $ {\varepsilon _0} $ , any formula containing $ {\varepsilon _0} $ can be utilized in as much as the units of the other variables are well known. For example, Using Gauss’s law, I can obtain the unit of $ {\varepsilon _0} $ as follows:
, we can drop the integral signs and vector symbols since they don’t affect our unit. Hence, we have
$\Rightarrow E \times A = \dfrac{Q}{{{\varepsilon _0}}} $
$\Rightarrow {\varepsilon _0} = \dfrac{Q}{{EA}} $
The unit of $ {\varepsilon _0} $ again can then be calculated as
$\Rightarrow {\varepsilon _{0U}} = \dfrac{C}{{\dfrac{N}{C} \times {m^2}}} = {C^2}{N^{ - 1}}{m^{ - 2}} $ which is an identical expression as the one calculated in the solution step. It is best to directly use expressions whose variables have SI units contained in the options.