Chapter 21 Problem 20

Problem

In the small angle approximation:

• the spheres only have horizontal displacement, and so the electric force of repulsion is always horizontal.
• $\tan \theta \approx \sin \theta \approx \theta$

Since the spheres are in equilibrium, the net force in each direction is zero.

(a)

$\sum F_{1x}=F_{T1}\sin \theta _{1}-F_{E1}=0\rightarrow F_{E1}=F_{T1}\sin \theta _{1}$

$\sum F_{1y}=F_{T1}\cos \theta _{1}-m_{1}g=0\rightarrow F_{T1}={\frac {m_{1}g}{\cos \theta _{1}}}\rightarrow F_{E1}={\frac {m_{1}g}{\cos \theta _{1}}}\sin \theta _{1}=m_{1}g\tan \theta _{1}=m_{1}g\theta _{1}$

similarly

$F_{E2}=m_{2}g\theta _{2}$

Apply Newton's third law:

$F_{E1}=F_{E2}\rightarrow m_{1}g\theta _{1}=m_{2}g\theta _{2}\rightarrow \theta _{1}{\big /}\theta _{2}=m_{2}{\big /}m1$

Thus the answer is 1

(b)

$\theta _{1}{\big /}\theta _{2}=m_{2}{\big /}m1=2$

(c)

The distance between the two spheres in small angle approximation is

$d=l\sin \theta _{1}+l\sin \theta _{2}\approx (\theta _{1}+\theta _{2})$

in the first case $\theta _{1}=\theta _{2}$ thus:

$d=2l\theta _{1}\rightarrow \theta _{1}={\frac {d}{2l}}$

$m_{1}g\theta _{1}=F_{E1}={\frac {kQ(2Q)}{d^{2}}}=mg{\frac {d}{2l}}$

$d=\left({\frac {4lkQ^{2}}{mg}}\right)$

in the second case $2\theta _{1}=\theta _{2}$ thus:

$d={\frac {3}{2}}l\theta _{1}\rightarrow \theta _{1}={\frac {2d}{3l}}$

$m_{1}g\theta _{1}=F_{E1}={\frac {kQ(2Q)}{d^{2}}}=mg{\frac {2d}{3l}}$

$d=\left({\frac {3lkQ^{2}}{mg}}\right)$