Question

[L] = [c]^a [G]^b ${\left[\frac{e^2}{4{\mathrm{\pi \epsilon }}_0}\right]}^c$
[L] = [LT^-–1]a [M–^-1L³T^-–2]^b [ML³T^-2]^c
[L] = L^a+3b+3c M^-–b+c T–^{-a-–2b-–2c}
a + 3b + 3c =1
- –b + c = 0
a + 2b + 2c = 0
On solving,
a = –- 2 , b = $\frac{1}{2} , c = \frac{{\displaystyle 1}}{{\displaystyle 2}}$
$\therefore$ L = $\frac{1}{c^2}{\left[G.\frac{e^2}{4\pi {\epsilon }_0}\right]}^{\frac{1}{2}}$