Physics, asked by TanujRao3795, 11 months ago


A physical quantity of the dimensions of length that can be formed out of c, G and e^2/4π£o an is (c is velocity of light, G is universal constant of gravitation and e is charge).write that physical quantity?

Answers

Answered by goblin19
3
epsilon, c is velocity of light , G is universal constant of gravitation and e is charge }. Terms and conditions : explanation ...
Answered by Anonymous
3

Question:

A physical quantity of the dimensions of length that can be formed out of c, G and \sf \dfrac{ {e}^{2} }{4\pi \varepsilon_{0}} is.

[c is velocity of light, G is the universal constant of gravitation and e is charge]

Solution:

Let's suppose the physical quantity formed by c, G and \sf \dfrac{ {e}^{2} }{4\pi \varepsilon_{0}} is  [c]^a[G] ^b \bigg[ \dfrac{ {e}^{2} }{4\pi \varepsilon_{0}}\bigg]^c

According to the question, it is a physical quantity for length.

So,

L= [c]^a[G] ^b \bigg[ \dfrac{ {e}^{2} }{4\pi \varepsilon_{0}}\bigg]^c  

Use dimensional analysis,

Dimensions of the following are:

  • L= [L]
  •  c=[LT^-1]
  •  G=[M^{-1}L^3T{-2}]
  •  \dfrac{ {e}^{2} }{4\pi \varepsilon_{0}} =[ML^3T^{-2}]

Why   \dfrac{ {e}^{2} }{4\pi \varepsilon_{0}} =[ML^3T^{-2}]?

Use dimensional analysis for each quantity in  \dfrac{ {e}^{2} }{4\pi \varepsilon_{0}}

Here we have

 e^2=[q] ^2=[AT]^2

 \epsilon_0=[M^{-1}L^{-3}T^4A^2]

Rest are Constants

Combining these two according to the formula we get,

 \dfrac{ {e}^{2} }{4\pi \varepsilon_{0}}=\frac{[AT]^2}{[M^{-1}L^{-3}T^4A^2]}=\frac{1}{[M^{-1}L^{-3}T^2]} = [ML^3T^{-2}]

Now, equating the dimensional formulas for

L= [c]^a[G] ^b \bigg[ \dfrac{ {e}^{2} }{4\pi \varepsilon_{0}}\bigg]^c \\\\ L=[LT^-1]^a[M^{-1}L^3T{-2}]^b[ML^3T^{-2}] ^c \\\\L=[L^aT^{-a}] [M^{-b} L^{3b}T^{-2b}][M^cL^{3c}T^{-2c}] \\\\ L=[L^{a+3b+3c}][M^{-b+c}][T^{-a-2b-2c}]

Now,equating the exponent power we have,

a+3b+3c=1 - - - - [i]\\\\ -b+c=0 - - - - [ii] \\\\ - a-2b-2c=0 - - - - [iii] \\\\ From [ii], we~have~b=c,~using~this~value~in~[iii] ~we, ~get \\\\ - a-2c-2c=0 \\\\ - a-4c=0 a=-4c \\\\ Using~a=-4c~and~b=c~in~[i]~we~have~\\\\ - 4c+3c+3c=1 \\\\ 2c=1 \\\\ c=\frac{1}{2} =b \\\\ We~ know~that~a=-4c ~so, ~a=-4 \times \frac{1}{2} ~, a=-2  

The physical quantity so obtained is

L= [c]^-2[G] ^{\frac{1}{2}}\bigg[ \dfrac{ {e}^{2} }{4\pi \varepsilon_{0}}\bigg]^{\frac{1}{2}} \\\\ => \frac{1}{c^2} G^{\frac{1}{2}}\bigg[ \dfrac{ {e}^{2} }{4\pi \varepsilon_{0}}\bigg]^{\frac{1}{2}} \\\\ =\frac{1}{c^2}  \bigg[G \frac{e^{2} }{4\pi \varepsilon_{0}} \bigg]^{\frac{1}{2}}

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