Two small conductiong ball of identical mass 20 g and identical charge `10^(-10) C` hang from non-conducting threads of length, L = 300 cm. If the equilibrium separation of balls is x and `x lt lt L` then the magnitude of x is (Assume, `4pi in_(0) =1/(9xx10^(9))` F/m and `g=10 m//s^(2)`)

Options

A. `2/5^(1//3)` mm
B. `3/10^(1//3)` mm
C. `3^(1//3)/10` mm
D. `3^(2//3)/5` mm

Answer

B

Explanation

Given, mass of each spherical ball,
`m=20g=2xx10^(-2) kg, g=10 m//s^(2)`
charge on each spherical ball,
`q_(1)=q_(2)=q=10^(-10) C`
and length of thread,
`L=300 cm=3 m`
According to the question,

Equilibrium at point B,
`T cos theta =mg` ...(i)
`T sin theta=F`
`T sin theta=1/(4pi epsi_(0)).q^(2)/x^(2)` ...(ii)
(From Coulomb's law)
From Eqn. (i) and (ii), we get
`(T sin theta)/(T cos theta)=(1/(4pi epsi_(0)).q^(2)/x^(2))/(mg)`
`["Given", 4pi epsi_(0)=1/(9xx10^(9))C^(2)//N-m^(2)]`
`:. mg tan theta =9xx10^(9).q^(2)/x^(2)` ...(iii)
From `Delta ABM`,
`tan theta=(x/2)/sqrt(L^(2)-(x/2)^(2))=x/sqrt(4L^(2)-x^(2))=x/(2L)" "( :' L gt gt x)`
`:.` From Eq. (iii), we get
`mg. x/(2L)=(9xx10^(9)xxq^(2))/x^(2)`
Putting the given values, we get
`2xx10^(-2)xx10xx x/(2xx3)=(9xx10^(9)xx10^(-20))/x^(2)`
`implies x^(3)=27/10^(10) m`
`implies x=3/(10^(3)xx10^(1//3))=3/10^(1//3)` mm
Therefore, magnitude of equilibrium separation distance of ball is `3/10^(1//3)` mm.