Question #54e33

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Question #54e33

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Answer 1 · 2016-12-06T08:12:14

We draw a free-body diagram for one of the charged balls, say left. This is shown in the figure below.
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All the forces acting on the charged ball are in equilibrium.

One force is tension $T$ $T$ in the silk thread. The other force is weight $m g$ $mg$ of ball acting downward and third is electrostatic force of repulsion $F_{elec}$ $F_"elec"$ between the two charged balls.

Equating the magnitude of vertical component of tension with weight of the ball we get
$T cos θ = m g$ $Tcostheta=mg$ .....(1)
and equating the magnitude of horizontal component of tension with the magnitude of electrostatic force we get
$T sin θ = F_{elec} = k_{e} \frac{q^{2}}{x^{2}}$ $Tsintheta=F_"elec"=k_eq^2/x^2$ .....(2)
where $x$ $x$ is the distance between the two charged balls.

Dividing (2) by (1) we get
$tan θ = k_{e} \frac{q^{2}}{x^{2}} / m g$ $tan theta=k_eq^2/x^2/ mg$ ....(3)

It is given that the angle between the silk threads is small ( $2 θ$ $2theta$ in our figure). We know that for small angles $θ$ $theta$

$tan θ \approx sin θ$ $tanthetaapproxsintheta$ ....(4)

From the geometry of the figure we see that
$sin θ = \frac{\frac{x}{2}}{l} = \frac{x}{2 l}$ $sin theta=(x/2)/l=x/(2l)$ .....(5)
Using (4) and (5) equation (3) reduces to
$\frac{x}{2 l} \approx k_{e} \frac{q^{2}}{x^{2} m g}$ $x/(2l)~~k_eq^2/(x^2 mg)$

Substituting the value of Coulomb's constant $k_{e}$ $k_e$ in terms of electric permittivity of free space $ε_{\circ}$ $epsilon_@$ and solving for $x$ $x$ we get
$x^{3} = \frac{1}{4 π ε_{\circ}} \times 2 l \frac{q^{2}}{m g}$ $x^3=1/(4pi epsilon_@)xx2lq^2/( mg)$
$\Rightarrow x = {(\frac{q^{2} l}{2 π ε_{\circ} m g})}^{\frac{1}{3}}$ $=>x=((q^2l)/(2pi epsilon_@mg))^(1/3)$

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Question #54e33

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