The force between two charges q1 and q2 separated by a distance d in a vacuum is given by F = q1q2/4πε0 d², where ε0 is a constant whose dimensions are [M L^-3 T^4 A²]. Check the dimensional balance of this equation.

Question

The force between two charges [latex]q_{1} and q_{2}[/latex] separated by a distance d in a vacuum is given by [latex]F = q_{1}q_{2}/4πε_{o}d^{2}[/latex], where e0 is a constant whose dimensions are [latex][M L^{-3} T^{4} A^{2}][/latex]. Check the dimensional balance of this equation.

Accepted Answer

The left-hand side of the equation is simply the force F and from Example 1.1 (2) we see that its dimensions are [latex][M L T^{-2}][/latex].
The dimensions of the right-hand side are [latex][q][q]/[4][π][ε_{0}][d^{2}][/latex]. From Example 1.3 (1) we see that the dimensions of electric charge [q] are [A T]. Numbers are dimensionless so that the figure 4 and the constant π have no dimensions. We are told in the question that the dimensions of [latex]ε_{0}[/latex] are [latex][M^{-1} L^{-3} T^{4} A^{2}][/latex]. The distance between the charges, d, has the dimensions of length so that [latex]d^{2}[/latex] has the dimensions [latex][L^{2}][/latex]. The dimensions of the right-hand side of the equation are therefore

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

which is the same as that obtained for the left-hand side of the equation. The equation is therefore dimensionally balanced.

Question 1.5: The force between two charges q1 and q2 separated by a dista...

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