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Let [epsilon(0)] denote the dimensional ...

Let `[epsilon_(0)]` denote the dimensional formula of the permittivity of vacuum. If M = mass, L = length, T = time and A = electric current, then:

A

`[epsilon_0] = M^(-1)L^3 T^2A`

B

`[epsilon_0]=M^(-1)L^(-3)T^4 A^2`

C

`[epsilon_0]= MLT^(-2)A^(-2)`

D

`[epsilon_0]=ML^2T^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
B
By Coulomb law,
`F=(1)/(4pi epsilon_0)(q_1 q_2)/(R^2)`
`implies epsilon_0 = (q_1 q_2)/(4pi FR^2)`
Substituting the units in the equation,
`epsilon_0 =(C^2)/(N . m^2)`
`=([AT]^2)/(MLT^(-2)L^2) =M^(-1)L^(-3)T^4 A^2`
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