Meandering charge

First off, it is clear that the original resultant force of the first 2 charges on the third charge is acting only downwards, since the horizontal forces cancel out. Qualitatively, it is obvious that the third charge will be attracted downwards by the first 2 charges, then because of its inertia, it would be able to continue downwards till $y = -0.01$ , before being attracted by the first 2 charges again and repeating the process to reach $y = 0.01$ again. Now we consider a quantitative description.

Let the third charge be at a position $y = x$ . The distance between this charge and either of the other 2 charges is $\sqrt{4 + x^2}$ by Pythagoras' Theorem. The force acting on the third charge by one of the first 2 charges is given by the formula, $F = k \frac{q_1 q_2}{r^2} = (9 \times 10^9) \frac{(10^{-3})(10^{-4})}{(4 + x^2)} = \frac{900}{4+x^2}$ . Now, since the horizontal forces will clearly cancel out, we are only concerned with the vertical forces, hence, we calculate the vertical component of this force due both of the first 2 charges: $F_{y,eff} = 2(\frac{900}{4+x^2})(\frac{x}{ \sqrt{4 + x^2}} ) = \frac{1800x}{\sqrt{(4 + x^2)^3}}$ .

In this case, $x$ is small, since $x$ ranges from $-0.01$ to $0.01$ only. Hence, we can approximate $\sqrt{4 + x^2} \approx \sqrt{4} = 2$ .

Thus, we have that $F_{y,eff} \approx ( \frac{1800}{8} ) x = 225x$ . If we consider the direction of the force in relation to the displacement $x$ , it is obvious that the force acts in the opposite direction to the displacement, and hence this is the familiar form of Simple Harmonic Motion $F = -kx$ . In this case, $k = 225$ , and we have that the period of SHM is as such: $T = \frac{2 \pi}{ \omega} = \frac{ 2 \pi }{ \sqrt{ \frac{k}{m} }} = \frac{ 2 \pi }{ \sqrt{ \frac{225}{10^{-3}} } } = 0.0132$ .

Sacrificing 3 orders of magnitude, assume the distance from the third charge to each of the other charges is always 2m. Then each of the positive charges, by Coulomb's law, exerts a force of $\frac{kQ_1Q_2}{r^2} = -225N$ on the negative charge with a vertical component of $\frac{-225y}{2} N$ . The horizontal components cancel out, so the negative charge experiences a net force of about -225y N along the y-axis, which we notice follows Hooke's law with k = 225. Thus, we have simple harmonic motion with period $2\pi \sqrt{\frac{m}{k}} = \boxed{.0132 s}$ .

Let $\theta$ be the angle between the moving charge and the $x$ -axis charge, measured from either of the fixed charges. Since $y$ is so small compared to $x$ , we make the simplifying approximation that $\theta \approx \sin \theta \approx \tan \theta$ and $\cos \theta \approx 1$ . Thus $y = 2 \tan\theta \approx 2 \theta$ .

The force exerted by a fixed charge on the moving charge is $k q_1 q_2 / (4 \cos^2 \theta) \approx -225$ Newtons. The vertical component of this force is $- 225 \sin \theta \approx -225 \theta$ , so the two charges exert a combined force of $-450 \theta$ (the horizontal components cancel out). The acceleration of the particle is therefore $\ddot{y} = -450000 \theta \approx -225000 y$ .

This is an equation of simple harmonic motion with period $2 \pi / \sqrt{225000} = \pi /75\sqrt{10} = 0.013246$ seconds.

According to coloumb law the return force on the negative charge is 2xAbs{9x10^9x(1E-3x-0.1E-3)/2^2}=450N.

The return force acting on the negative charge along the displacement line is 450x(0.01/2)=2.25N, but this force is also displacing the charge 0.01 meter. So let's see this situation as of an oscillating spring with a K constant of K= 225/0.01=225N/m.

The period of which is given by T=2xPIxSQRT(M/K) where in this case we have M=0.001Kgmass and K=225N/M. Crank the formula for T using the last data and that will give T=0.01324 Sec

Net force on 0.1mC charge is given as –(2kqQx)/(2^2 + x^2)^3/2

at a distance x from origin

as x = 0.01 <<< 4 so neglecting x in denominator, we get force = –(2kqQx)/(2^2)^3/2

comparing it with equation of simple harmonic motion, we get

w^2 = (2kqQ)/m(2^2)^3/2 w= angular frequency

and time taken to return back is = time period of SHM

                                                   = 2π/w

                                                    = 2π x 4(m/kqQ)^1/2

                 Putting values given we get T = 0.0133s

$F_{1,x}=-F_{2,x}\Rightarrow F_{net,x}=0\\ F_{1,y}=F_{2,y}=\frac{1}{4\pi \epsilon }\frac{Qq}{r^2}\left(\frac{y}{r}\right)\\\Rightarrow F_{net}=F_{1,y}+F_{2,y}=2\left(\frac{1}{4\pi \epsilon }\frac{Qqy}{r^3}\right)\\ r=\sqrt{d^2+y^2}\approx d, \text{since } d=2\text{m}\gg y=0.01\text{m}\\ F_{net}=2\left(\frac{1}{4\pi \epsilon }\frac{Qqy}{d^3}\right)=-ky\\ k=-2\left(9\times10^9\times\frac{1\times10^-3\times(-0.1)\times10^-3}{2^3}\right)=225 \text{ N/m}\\ T=2\pi \sqrt{\frac{m}{k}}=2\pi\sqrt{\frac{1\times10^-3}{225}}=0.0132 \text{ sec.}$

To find the motion of the charge q we first need to find the forces acting on it. Since the problem is symmetrical the force in the x-direction cancels out. We are therefore left with the y-component of the force, which is

$F_y=-\frac{k Q q}{(-2)^2+y^2} \frac{y} {((-2)^2+y^2)^{1/2}} -\frac{k Q q}{2^2+y^2} \frac{y} {(2^2+y^2)^{1/2}}=-\frac{2k Q qy}{(4+y^2)^{3/2}}$ ,

from Coulomb's law and the geometry that projects out the y-component. y in the above expression is the y coordinate of q. Since y<<2, we can neglect $y^2$ in the denominator and we see something interesting - the force is exactly the same form as a simple harmonic oscillator $F_y=-Ky$ where $K$ is a constant! Therefore the motion is simple harmonic motion and we can read the period off as

$T=2\pi \sqrt{\frac {m} {K}}=2 \pi \sqrt{\frac{4m}{kQq}}=0.013~s$ .