Solveeit Logo

Question

Physics Question on Dimensional Analysis

Find the dimensions of electric permittivity.

A

[A2M1L3T4][A^2M^{-1}L^{-3}T^4]

B

[A2M1L3T0][A^2M^{-1}L^{-3}T^0]

C

[AM1L3T4][AM^{-1}L^{-3}T^4]

D

[A2M0L3T4][A^2M^{0}L^{-3}T^4]

Answer

[A2M1L3T4][A^2M^{-1}L^{-3}T^4]

Explanation

Solution

From Coulomb?? law, the force of attraction/repulsion between two point charges qq, qq separated by distance rr is
F=14πε0q2r2F = \frac{1}{4\,\pi\,\varepsilon_{0}} \frac{q^{2}}{r^{2}}
ε0=14πq2Fr2\Rightarrow \varepsilon _{0} = \frac{1}{4\,\pi } \cdot \frac{q^{2}}{Fr^{2}}
where e0e_{0} is electric permittivity.
Dimensions of ε0=[AT]2[MLT2][L2]\varepsilon_{0} =\frac{\left[AT\right]^{2}}{\left[MLT^{-2}\right]\left[L^{2}\right]}
ε0=[A2M1L3T4]\varepsilon_{0} = \left[A^{2}\,M^{-1}\,L^{-3}\,T^{4}\right]