Physics

The dimension of permittivity of vacuum 


ANSWER

$$[M^{ -1} L^ {-3} T^4 A^2]$$


SOLUTION
The permittivity of the vacuum can be given by:
$$F_e=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q^2}{r^2}$$
$$permitivity\ (\varepsilon_0)= \dfrac{(q)^2}{(F_e)\times(r)^2}$$

$$=\dfrac{[TA]^2}{[MLT^ {-2}] [L]^2}$$

$$=\dfrac{[TA]^2}{[ML^3T^ {-2}]}$$

$$= [M^{-1} L^{-3} T^4 A^2]$$

$$= [M^{ -1} L^ {-3} T^4 A^2]$$
This is the required solution.
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