Question

The dimensions of $\varepsilon_0$ (permittivity in free space) is

• A. $ML^2T^4A^2$
• B. $ML^{-3}T^2A^2$
• C. $M^{-1}L^3T^4A^2$
• D. $ML^3T^2A^2$
• E. $M^{-1}L^{-3}T^4A^2$

Hint:

If you forget Coulomb's law, you can also use the capacitance formula $C = \frac{\varepsilon_0 A}{d}$ to find dimensions, where $C = Q/V$.

Answer 1

Answer: (E)

Concept:
We can derive the dimensions of $\varepsilon_0$ using Coulomb's Law: $F = \frac{1}{4\pi\varepsilon_0} \frac{q_1q_2}{r^2}$. Rearranging for $\varepsilon_0$: \[ \varepsilon_0 = \frac{q_1q_2}{4\pi F r^2} \]

Step 1: Substitute fundamental dimensions.
[itemsep=4pt]
• Charge ($q$): $Current \times Time = [AT]$
• Force ($F$): $[MLT^{-2}]$
• Distance ($r$): $[L]$

Step 2: Solve the dimensional equation.
\[ [\varepsilon_0] = \frac{[AT][AT]}{[MLT^{-2}][L^2]} = \frac{[A^2T^2]}{[ML^3T^{-2}]} = [M^{-1}L^{-3}T^4A^2] \]