Question

A physical quantity of the dimension of length that can be formed out of c, G and e^2/4πϵ0 is [c is velocity of light, G is universal constant of gravitation and e is charge.

Answer 1

Answer:

$[L]=[c]^{-2}[G]^{1\over2}{[{e^2\over4\pi\epsilon_{o}}]^{1\over2}}$

Explanation:

Dimensions of given quantities are-

$[c]=[LT^{-1}]\\\\\[[G]=[M^{-1}L^3T^{-2}]\\\\\[[{e^2\over4\pi\epsilon_{o}}]=[ML^3T^{-2}]$

Hence, for a physical quantity of dimension of length,

$[M^0L^1T^0]=[c]^a[G]^b{[{e^2\over4\pi\epsilon_{o}}]^c}\\\\\[[M^0L^1T^0]=[LT^{-1}]^a[M^{-1}L^3T^{-2}]^b[ML^3T^{-2}]^c\\\\\[[M^0L^1T^0]=[M^{-b+c}L^{a+3b+3c}T^{-a-2b-2c}]$

Hence, equating powers, we get

0 = -b + c        ⇒ b = c       .................(1)

1 = a + 3b + 3c   ⇒ 1 = a + 3b + 3b  ⇒ 1 - a = 6b     ............(2)

0 = -a -2b -2c  ⇒ a = -2b - 2b  ⇒ a = -4b   ...............(3)

From (2),

1 - a = 6b

1 - (-4b) = 6b       ..........(from (2))

b = $1\over2$

∴ b = c = $1\over2$     (from(1))

∴ a = -4b = -2       (from (3))

Putting the value of a, b and c

$[M^0L^1T^0]=[c]^{-2}[G]^{1\over2}{[{e^2\over4\pi\epsilon_{o}}]^{1\over2}}\\\\\[[L]=[c]^{-2}[G]^{1\over2}{[{e^2\over4\pi\epsilon_{o}}]^{1\over2}}$

Answer 2

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

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