A physical quantity of the dimensions of length that can be formed out of c, G and e^2/4πε0

Question

A physical quantity of the dimensions of length that can be formed out of c, G and \(\frac{e^2}{4\pi \epsilon_0}\) is [c is velocity of light, G is universal constant of gravitation and e is charge]:

(A) \(\frac{1}{c^2}[G\frac{e^2}{4\pi \epsilon_0}]\)

(B) \(c^2[G\frac{e^2}{4\pi \epsilon_0}]^{1/2}\)

(c) \(\frac{1}{c^2}[\frac{e^2}{G4\pi \epsilon_0}]^{1/2}\)

(D) \(\frac{1}{c}G\frac{e^2}{4\pi \epsilon_0}\)

Answer 1

(A) \(\frac{1}{c^2}[G\frac{e^2}{4\pi \epsilon_0}]^{1/2}\)

Let the physical quantity formed of the dimensions of length be given as.

[L] = [c]^x [G]^y \([\frac{e^2}{4\pi \epsilon_0}]^z\) …………….. (i)

Now,

Dimensions of velocity of light [c]^x = [LT^-1]^x Dimensions of universal gravitational constant [G]^y = [L³T²M^-1]^y

Dimensions of \([\frac{e^2}{4\pi \epsilon_0}]^z\) = [ML³T^-2]^z

Substitrning these in equation (i)

[L] [LT^-1]^x [M^-1L³T^-2]y [ML³T^-2]^z

= L^x+3y+3z M^-y+z T^-x-2y-2z

Solving for x, y, z

x + 3y + 3z = 1

-y + z = 0

x + 2y + 2z = O

Solving the above equation,

x = -2, y = \(\frac{1}{2}\), z = \(\frac{1}{2}\)

∴ L = \(\frac{1}{c^2}\)\([G\frac{e^2}{4\pi \epsilon_0}]^{1/2}\)