Question

Suppose [

e

0

] is permittivity of free space. If M = mass, L = length, T = time and A = electric

current, then .​

Answer 1

Explanation:

Dimension formula of ε

0

ε

0

=

4πF

1

r

2

v

1

v

2

(F=MLT

−2

)

ε

0

=

MLT

−2

1

l

2

LAT×AT

=M

−1

L

−3

t

4

A

2

Dimension of μ

0

N/A

2

or WbA

−1

m

−1

=[M][L][T]

−2

[A]

−2

Answer 2

here is ur answer:---

Explanation:

**Given:

e = permittivity of free space

M = mass

L = length

T = time

A = current

electrostatic force ,

F = (1/4πe) (q1q2/ r^2)

dimension of force =

$m \: l \: {t}^{ - 2}$

✓✓ the numerical value and π are dimension less constants

q = charge = current× time

dimension of q = [ AT ]

dimension of r^2 = [ L^2 ]

$m \: l \: {t}^{ - 2} \: = \: ( \frac{1}{e} ) \times \frac{(at)(at)}{ {l}^{2} }$

$e \: = \: \frac{( {a}^{2} \times {t}^{2}) }{(m \: l \: {t}^{ - 2}) \times ( {l}^{2} )}$

$e \: = \: {m}^{ - 1} {l}^{ - 3} {t}^{4} {a}^{2}$

hope it helps u