In electrostatics, Coulomb's law states that the force F between two point charges q_1 and… — JAMB UTME

In electrostatics, Coulomb's law states that the force $F$ between two point charges $q_1$ and $q_2$ separated by a distance $r$ in a vacuum is given by $F = \frac{q_1 q_2}{4\pi \varepsilon_0 r^2}$ , where $\varepsilon_0$ is the permittivity of free space. What are the fundamental dimensions of $\varepsilon_0$ expressed in terms of mass ( $M$ ), length ( $L$ ), time ( $T$ ), and electric current ( $I$ )?

$M^{-1} L^{-3} T^4 I^2$ Answer
B
$M^{-1} L^{-3} T^2 I^2$
C
$M L^3 T^{-4} I^{-2}$
D
$M^{-1} L^{-2} T^4 I^2$

Answer

The dimensions of permittivity of free space

\varepsilon_0

are

M^{-1} L^{-3} T^4 I^2

The correct answer is derived by isolating

\varepsilon_0 = \frac{q_1 q_2}{4\pi F r^2}

. Substituting

[q] = I T

[F] = M L T^{-2}

, and

[r] = L

gives

[\varepsilon_0] = \frac{I^2 T^2}{M L^3 T^{-2}} = M^{-1} L^{-3} T^4 I^2

Step-by-Step Solution

Rearrange Coulomb's Law to isolate permittivity of free space

\varepsilon_0

\varepsilon_0 = \frac{q_1 q_2}{4\pi F r^2}

Isolating

\varepsilon_0

allows us to substitute the dimensions of each constituent physical quantity.

Determine the dimensions of charge

q

, force

F

, distance

r

, and the constant

4\pi

[q] = I T

[F] = M L T^{-2}

[r^2] = L^2

, and

[4\pi] = 1

(dimensionless)

Electric current is a base unit (

I

), so electric charge is current multiplied by time (

I T

). Force is mass times acceleration (

M L T^{-2}

Substitute the fundamental dimensions into the rearranged equation and simplify exponent powers

[\varepsilon_0] = \frac{(I T)(I T)}{(M L T^{-2})(L^2)} = \frac{I^2 T^2}{M L^3 T^{-2}} = M^{-1} L^{-3} T^4 I^2

Applying exponent laws:

T^2 / T^{-2} = T^{2 - (-2)} = T^4

1 / M = M^{-1}

, and

1 / L^3 = L^{-3}

Key Concept

Dimensional analysis of physical constants in electromagnetism

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