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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 by Expert (11.6k points) | 37 views

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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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