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\prob{83}
9277_83

Two pith balls of equal mass M and equal charge q are suspended from the same point on long massless threads of length L as shown in the figure above. If k is the Coulomb's law constant, then for small values of $\theta$, the distance d between the charged pith balls at equilibrium is


  1. $\left(\frac{2kq^2L}{Mg}\right)^{1/3}$
  2. $\left(\frac{kq^2L}{Mg}\right)^{1/3}$
  3. $\left(\frac{2kq^2L}{Mg}\right)^{1/2}$
  4. $\left(\frac{kq^2L}{Mg}\right)^{1/3}$
  5. $L/4$

Electromagnetism}Forces

Sum of the forces for one of the mass in the x (horizontal) and y (vertical) directions gives,

For small angles, \cos\theta \approx 1 \Rightarrow T \approx mg. From the geometry, one can deduce that \sin\theta = (d/2)/L.

Thus, the x equation yields T(d/2)/L=kq^2/d^2 \Rightarrow d^3 = 2kq^2L/(mg) (since T\approx mg from the y equation for small angles). This is choice (A).

See below for user comments and alternate solutions! See below for user comments and alternate solutions!
Alternate Solutions
cjohnson415
2013-08-19 01:40:11
Here's an alternative solution that skips the trig altogether:

Due to small \theta, we can consider each as a simple pendulum, a special case of the simple harmonic oscillator:

\omega = \sqrt{\frac{k}{m}} = \frac{2 \pi} {T} = \sqrt{\frac{g}{L}}

Thus k = \frac{mg}{L}, which is the restoring force constant (F = -kx), so:

\frac{mgd}{2L} = \frac{k_{c}q^2}{d^2}

The rest is the same algebra, just a different/potentially quicker/potentially prettier way to arrive at the force balance.
Alternate Solution - Unverified
neo55378008
2012-08-18 12:27:15
Dimensional analysis helps!
Since \\F=\frac{kq^2}{r^2} is in N, \\k q^2 will have units of \\N m^2
Mg has units of N
so \frac{kq^2L}{Mg} will have units of \\m^3
This leaves only A and B. Knowing that solving for \theta will give half of d, take the solution with an extra 2 in it! (A)
Alternate Solution - Unverified
redmomatt
2011-11-06 08:25:21
Heres another take without the trig functions...

Since \theta is small, we can remove all of the trig functions. Thus, we just need to equate the outward Coulomb force F_c to the force of gravity along the horizontal F_x.

\theta = \frac{F_x}{mg} \Rightarrow F_x = mg\theta = \frac{mgd}{2L}, since L\theta = d/2.

Hence,

F_c = F_x \Rightarrow \frac{ k q^2 }{d^2} = \frac{mgd}{2L} \Rightarrow d^3 = \frac{2kq^2L}{mg} \Rightarrow d = \left(\frac{2kq^2L}{mg}\right)^{1/3}, which is (B) .
Alternate Solution - Unverified
Andresito
2006-03-29 14:56:46
It is a bit faster to equate F(electric) = T*Sin(theta), m*g = T*Cos(theta) and divide them. Tan(theta) ~ thetaAlternate Solution - Unverified
Comments
astropolo
2014-07-29 11:55:16
I have a quick question about the small angle approximations. Why can you approximate that Cos~1 while you use the geometry for Sin rather than saying sin~theta?
JoshWasHere
2014-08-20 08:20:22
When you expand cosine and sine, you see that the first two terms in the cosine and sine expansion are cos\theta = 1 - \frac{\theta ^2}{2!} and sin\theta = \theta - \frac{\theta ^3}{3!}. Because of the small angle, we select only the first order terms, and drop all higher order terms (with the condition that \theta < 5 degrees).
Answered Question!
cjohnson415
2013-08-19 01:40:11
Here's an alternative solution that skips the trig altogether:

Due to small \theta, we can consider each as a simple pendulum, a special case of the simple harmonic oscillator:

\omega = \sqrt{\frac{k}{m}} = \frac{2 \pi} {T} = \sqrt{\frac{g}{L}}

Thus k = \frac{mg}{L}, which is the restoring force constant (F = -kx), so:

\frac{mgd}{2L} = \frac{k_{c}q^2}{d^2}

The rest is the same algebra, just a different/potentially quicker/potentially prettier way to arrive at the force balance.
Alternate Solution - Unverified
greatspirits
2012-11-02 15:49:28
Energy conservation anyone? The change in gravitational potential energy should equate to potential between the charges. So... 2mg\delta h should equate to... hmmm well I usually prefer the energy method but the force way seems to be more effecient here.NEC
neo55378008
2012-08-18 12:27:15
Dimensional analysis helps!
Since \\F=\frac{kq^2}{r^2} is in N, \\k q^2 will have units of \\N m^2
Mg has units of N
so \frac{kq^2L}{Mg} will have units of \\m^3
This leaves only A and B. Knowing that solving for \theta will give half of d, take the solution with an extra 2 in it! (A)
henryb
2016-09-12 18:14:40
thx a lot. this is the simplest solution
Alternate Solution - Unverified
redmomatt
2011-11-06 08:25:21
Heres another take without the trig functions...

Since \theta is small, we can remove all of the trig functions. Thus, we just need to equate the outward Coulomb force F_c to the force of gravity along the horizontal F_x.

\theta = \frac{F_x}{mg} \Rightarrow F_x = mg\theta = \frac{mgd}{2L}, since L\theta = d/2.

Hence,

F_c = F_x \Rightarrow \frac{ k q^2 }{d^2} = \frac{mgd}{2L} \Rightarrow d^3 = \frac{2kq^2L}{mg} \Rightarrow d = \left(\frac{2kq^2L}{mg}\right)^{1/3}, which is (B) .
Alternate Solution - Unverified
Dodobird
2010-11-12 11:57:31
tan(theta) = d/2L
tan(theta) = kq*q/d*d*MG

d/2L = kq*q/d*d*MG

gives answer A.
NEC
FortranMan
2008-10-03 14:02:49
Also note because theta is so small here, we can approximate the vector component of the gravity that's NOT inducing tension to be d/2.NEC
JAS
2007-11-01 09:20:45
It seems easier to equate torques around the pivot point, where the torque on the right mass due to the repulsive electric force isrnrnL\frac{kq^2}{d^2}\cos\theta counterclockwisernrnand the torque on the right mass due to its weight is:rnrnMgL\sin\theta clockwisernrn\cos \theta \approx 1 and \sin \theta \approx \tan \theta = \frac{d}{2L}rnrnFrom this, choice A follows.NEC
relain
2007-10-23 19:21:51
typo in the question, option D should be a square not a cube root. It's currently identical to option B. NEC
Andresito
2006-03-29 14:56:46
It is a bit faster to equate F(electric) = T*Sin(theta), m*g = T*Cos(theta) and divide them. Tan(theta) ~ thetaAlternate Solution - Unverified

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