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

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

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

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

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

in Physics | 37 views

+1 vote

Correct option is B

From Coulomb's law we know,

F = $${1 \over {4\pi { \in _0}}}{{{q_1}{q_2}} \over {{r^2}}}$$

\therefore $${ \in _0} = {1 \over {4\pi }}{{{q_1}{q_2}} \over {F{r^2}}}$$

Hence, $$\left[ {{ \in _0}} \right] = {{\left[ {AT} \right]\left[ {AT} \right]} \over {\left[ {ML{T^{ - 2}}} \right]\left[ {{L^2}} \right]}}$$

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

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