Answer to Question #135384 in Mechanics | Relativity for Joe

Question

Two small spheres each suspended by a thread of small length, L from a common support have masses m and 2m respectively. Each sphere carries a net charge of +Q. Assuming that the angle the thread make with the vertical is small, and the distance, X between the spheres can be given by the equation X = (5KQ^2. L / 2mg)^1/3
Where g = acceleration due to gravity, K = 1 / 40 and 0 = permittivity of free space. Show that the equation is homogeneous.

Answer 1

Given equation is $X = (\frac{5KQ^2L}{2mg})^{1/3}$

where $K = \frac{1}{4\pi \epsilon_0}$

Equation will be homogeneous only when dimensions on both sides are same.

Since X is separation, so it dimension will be $[L]$

Dimension of K is $[ML^3T^{-4}A^{-2} ]$

Dimension of Q is $[AT]$

Dimension of L is [L]

Dimension of m is [M]

Dimension of g is $[LT^{-2}]$

Dimension of left hand side, $(\frac{[ML^3T^{-4}A^{-2}][A^2T^2][L] }{[M][LT^{-2}]})^{1/3} = [L]$

Since Dimension on both sides are same, so equation is homogeneous.