Question

Two similar conducting balls of mass m are hung from silk threads of length L and carry simillar charges q as shown in the figure. Assuming 0 to be small, the distance x between the ball is :

• A. ${\left[\frac{q^{2}L}{4\pi e_{o}mg}\right]}^{1/3}$
• B. ${\left[\frac{q^{2}L}{4\pi e_{o}mg}\right]}^{-1/3}$
• C. ${\left[\frac{q^{2}L}{4\pi e_{o}mg}\right]}^{3/2}$
• D. ${\left[\frac{q^{2}L}{4\pi e_{o}mg}\right]}^{1/2}$

Answer 1

The correct option is A ${\left[\frac{q^{2}L}{4\pi e_{o}mg}\right]}^{1/3}$

$T\sin \theta =Fe$

$T\cos \theta =mg$

${\left(\frac{x}{2}\right)}^2+b^2=L^2$

$b^2=L^2-{\left(\frac{x}{2}\right)}^2$

$b=\sqrt{L^2}=L$

$F_e=\frac{K.q.q}{x^2}$

$F_e=\frac{K.q^2}{X^2}⟶\left(1\right)$

$\frac{T\sin \theta }{T\cos \theta }=\frac{F_e}{mg}$

or, $\tan \theta =\frac{F_e}{mg}$

$mg\tan \theta =\frac{F_e}{mg}mg$

$mg\tan \theta =F_e$

$=\frac{{Kq}^2}{x^2}$ (putting equation $\left(1\right)$)

$mg\left(\frac{X}{2L}\right)=\frac{{Kq}^2}{x^2}{x}^2$

$\frac{X^{3}mg}{2L}=\frac{{Kq}^2}{1}$

$\frac{X^{3}mg}{mg}=\frac{{Kq}^2.2L}{1}$

$m{\left(x^3\right)}^{1/3}={\left(\frac{{Kq}^{2}L}{mg}\right)}^{1/3}$

$K=\frac{1}{4\pi {ϵ}_0}$ (replacing the value)

$x={\left(\frac{1}{4\pi {ϵ}_0}\frac{q^{2}L}{mg}\right)}^{1/3}$

$X={\left(\frac{1}{4\pi {ϵ}_0}\frac{q^{2}L}{mg}\right)}^{1/3}$