Two similar ball of mass m attached by silk thread of length a and carry similar charge q.assume θ is small enough that tanθ≈sinθ to this approximation,show that X=(((qa)/(2πε_0 mg)))^(1/3) where X is distance of separation.


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Question Number 50994 by peter frank last updated on 23/Dec/18

$T w o s i m i l a r b a l l o f m a s s$ $m a t t a c h e d b y s i l k t h r e a d$ $o f l e n g t h a a n d c a r r y$ $s i m i l a r c h a r g e q . a s s u m e θ i s$ $s m a l l e n o u g h t h a t$ $t a n θ \approx s i n θ t o t h i s$ $a p p r o x i m a t i o n, s h o w$ $t h a t X = {(\frac{q a}{2 π ε_{0} m g})}^{\frac{1}{3}}$ $w h e r e X i s d i s t a n c e o f$ $s e p a r a t i o n .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 23/Dec/18

	$T = t e n s i o n i n t h r e a d$ $T c o s θ = m g$ $T s i n θ = F$ $F = r e p u l s i v e f o r c e = \frac{1}{4 π ϵ_{0}} \frac{q^{2}}{x^{2}}$ $t a n θ = \frac{F}{m g}$ $s i n θ = \frac{F}{m g}$ $\frac{x}{2 a} = \frac{1}{4 π ϵ_{0}} \times \frac{q^{2}}{x^{2}} \times \frac{1}{m g}$ $x^{3} = \frac{{a q}^{2}}{2 π ϵ_{0}} \times \frac{1}{m g}$ $x = {(\frac{{a q}^{2}}{2 π ϵ_{0} m g})}^{\frac{1}{3}}$
	Answered by peter frank last updated on 23/Dec/18

	$F = \frac{q^{2}}{4 π ε_{0} r^{2}} [q_{1} = q_{2}]$ $\sin θ = \frac{x}{2 a}$ $\tan θ = \frac{F}{m g}$ $F = m g \tan θ$ $\tan θ \approx \sin θ = θ$ $F = m g \frac{x}{2 a} . . . . . . (i)$ $F = \frac{q^{2}}{4 π ε_{0} r^{2}} [r = x] . . . . . (i i)$ $m g \frac{x}{2 a} = \frac{q^{2}}{4 π ε_{0} x^{2}}$ $x = {[\frac{q^{2} a}{4 π ε_{0} m g}]}^{\frac{1}{3}}$
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