Answer:
[tex]\theta =\left (\frac{kq^{2}}{4L^{2}\times mg} \right )^{\frac{1}{3}}[/tex]
Explanation:
Let the length of the string is L.
Let T be the tension in the string.
Resolve the components of T.
As the charge q is in equilibrium.
T Sinθ = Fe ..... (1)
T Cosθ = mg .......(2)
Divide equation (1) by equation (2), we get
tan θ = Fe / mg
[tex]tan\theta =\frac{\frac{kq^{2}}{AB^{2}}}{mg}[/tex]
[tex]tan\theta =\frac{\frac{kq^{2}}{4L^{2}Sin^{\theta }}}}{mg}[/tex]
[tex]tan\theta =\frac{kq^{2}}{4L^{2}Sin^{2}\theta \times mg}[/tex]
[tex]tan\theta\times Sin^{2}\theta =\frac{kq^{2}}{4L^{2}\times mg}[/tex]
As θ is very small, so tanθ and Sinθ is equal to θ.
[tex]\theta ^{3} =\frac{kq^{2}}{4L^{2}\times mg}[/tex]
[tex]\theta =\left (\frac{kq^{2}}{4L^{2}\times mg} \right )^{\frac{1}{3}}[/tex]
