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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

`[in_(0]=[M^(-1)L^(-3)T^(2)A]`

B

`[in_(0)]=[M^(-1)L^(-3)T^(4)A^(2)]`

C

`[in_(0)]=[M^(-1)L^(2)T^(-1)A^(-2)]`

D

`[in_(0)]=[M^(-1)L^(2)T^(-1)A]`

Text Solution

Verified by Experts

The correct Answer is:
B
`F=(1)/(4pi in_(0))(q_(1)q_(2))/r^(2)rArr [MLT^(-2)]=[1/in_(0)] (A^(2)T^(2))/L^(2)`
`rArr [epsilon_(0)]=[M^(-1)L^(-3)T^(4)A^(2)]`
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