Question 209: Units , Dimensions and Error Analysis

Question

Let [$${\varepsilon _0}$$] denote the dimensional formula of the permittivity of vacuum. If M = mass, L = length, T = time
and A = electric current, then:

Accepted Answer

From Coulomb's law we know,
$$F = {1 \over {4\pi { \in _0}}}{{{q_1}{q_2}} \over {{r^2}}}$$
$$	herefore$$ $${ \in _0} = {1 \over {4\pi }}{{{q_1}{q_2}} \over {F{r^2}}}$$
Hence, $$\left[ {{ \in _0}} ight] = {{\left[ {AT} ight]\left[ {AT} ight]} \over {\left[ {ML{T^{ - 2}}} ight]\left[ {{L^2}} ight]}}$$
= $$\left[ {{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}} ight]$$

Answer

$${\varepsilon _0} = \left[ {{M^{ - 1}}{L^{ - 3}}{T^2}A} \right]$$

Answer

$${\varepsilon _0} = $$$$\left[ {{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}} \right]$$

Answer

$${\varepsilon _0} = \left[ {{M^1}{L^2}{T^1}{A^2}} \right]$$

Answer

$${\varepsilon _0} = \left[ {{M^1}{L^2}{T^1}A} \right]$$

Units , Dimensions and Error Analysis

Let [${\varepsilon _0}$] denote the dimensional formula of the permittivity of vacuum. If M = mass, L = length, T = time and A = electric current, then:

Choose the correct answer: