Dimensional Formula of Permittivity of Vacuum (ε₀)

To find the dimensional formula of ε₀, we use Coulomb's law for the force between two charges:

$F=\frac{1}{4\pi {\epsilon }_0}·\frac{q_1{q}_2}{r^2}$

Step-by-Step Derivation:

Step 1: Rearrange Coulomb's law to solve for ε₀:

${\epsilon }_0=\frac{1}{4\pi }·\frac{q_1{q}_2}{F{r}^2}$

Step 2: Since 4π is dimensionless, we ignore it for dimensional analysis:

$\left[{\epsilon }_0\right]=\frac{[q{]}^2}{\left[F\right][r{]}^2}$

Step 3: Substitute known dimensional formulas:

Charge [q] = [A T] (since current × time)

Force [F] = [M L T⁻²] (mass × acceleration)

Distance [r] = [L]

Step 4: Plug into the equation:

$\left[{\epsilon }_0\right]=\frac{{\left[AT\right]}^2}{\left[ML{T}^{-2}\right][L{]}^2}=\frac{A^2}{T^2}$

Step 5: Simplify the exponents:

$\left[{\epsilon }_0\right]=M^{-1}{L}^{-3}{T}^4{A}^2$

Final Answer: The correct dimensional formula is $\left[{\epsilon }_0\right]=\left[M^{-1}{L}^{-3}{T}^4{A}^2\right]$

Related Topics:

• Dimensional Analysis
• Coulomb's Law
• Electrostatics
• SI Units and Dimensions

Key Formulae:

Coulomb's Law: $F=\frac{1}{4\pi {\epsilon }_0}·\frac{q_1{q}_2}{r^2}$

Dimensional formula of force: [F] = [M L T⁻²]

Dimensional formula of charge: [q] = [A T]