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Physics
Physics
Problem
Let
[
ϵ
0
]
[\epsilon_0]
[
ϵ
0
]
denote the dimensional formula of the permittivity of vacuum. If
M
M
M
= mass,
L
L
L
= length,
T
T
T
= time and
A
A
A
= electric current, then: a.
ϵ
0
=
[
M
−
1
L
−
3
T
2
A
]
\epsilon_0 = [M^{-1}L^{-3}T^2A]
ϵ
0
=
[
M
−
1
L
−
3
T
2
A
]
b.
ϵ
0
=
[
M
−
1
L
−
3
T
4
A
2
]
\epsilon_0 = [M^{-1}L^{-3}T^4A^2]
ϵ
0
=
[
M
−
1
L
−
3
T
4
A
2
]
c.
ϵ
0
=
[
M
1
L
2
T
1
A
2
]
\epsilon_0 = [M^1L^2T^1A^2]
ϵ
0
=
[
M
1
L
2
T
1
A
2
]
d.
ϵ
0
=
[
M
1
L
2
T
1
A
]
\epsilon_0 = [M^1L^2T^1A]
ϵ
0
=
[
M
1
L
2
T
1
A
]
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