Q 1.3: Check that the ratio \frac{ke^{2}}{Gm_{e}m_{p}} is dimensionless. Look up a Table of Physical Constants and determine the value of this ratio. What does the ratio signify?

Answers (1)

Electrostatic force 

F=\frac{KQ^2}{r^2}

So the dimension of 

[Ke^2]=[Fr^2]..................(1)

The gravitational force between two bodies of mass M and m is

F=\frac{GMm}{r^2}

so dimension of 

[Gm_em_p]=[Fr^2].............(2)

Therefor from (1) and (2)

[\frac{ke^{2}}{Gm_{e}m_{p}} ]             is dimensionless

or

Here, 

K = 1/4\pi \epsilon _{0}, where  is the permittivity of space. [1/ \epsilon _{0}] =[C/V.m] = [Nm^2 C^{-2}]

e = Electric charge ([e] = [C])

G = Gravitational constant. ([G]= [Nm^2kg^{-2}] )

m_{e} and m_{p} are mass of electron and proton ([m_{e}] = [m_{p}]  = [Kg])

Substituting these units, we get

[\frac{ke^{2}}{Gm_{e}m_{p}} ] = [\frac{C^2 \times Nm^2C^{-2}}{Nm^2kg^{-2}\times kg\times kg} ] = M^{0}L^{0}T^{0}  

Hence, this ratio is Dimensionless.

 

Putting the value of the constants

\frac{ke^{2}}{Gm_{e}m_{p}} = \frac{9\times10^9\times(1.6\times10^{-19})^{2}}{6.67\times10^{11}\times9.1\times10^{-31}\times1.67\times10^{-27}}

=2.3 \times 10^{40}

 The given ratio is the ratio of electric force \frac{ke^{2}}{R^2}to the gravitational force between an electron and a proton \frac{Gm_{e}m_{p}}{R^2} considering the distance between them is constant!

 

 

 

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