Undergrad Electricity, Newtons Law question

Apply Newton’s law to the negative particle to find the nonlinear differential
equation governing the evolution of its y coordinate.



How does one begin attempting this problem?

Hello and welcome to TSR.

This is clearly part of a larger question and without that context it's impossible to answer you.

So post the full question including any diagrams you have and tell us your thoughts on how you might start solving it. Then hopefully someone on these boards will guide you through it.

Okey, thanks friend. =)

We are dealing with 3 charges;
2 positive, charge q which are a distance d apart in the x-direction.
Our negative charge of magnitude 4q; is at the x-coordinate half way between the two positive charges.

Is that the full question or is there other information this question is building on? As it stands, the question makes little sense, specifically; "the evolution of it's y co-ordinate".

First we must assume that the x and y co-ordinates are the reference system for the (2-dimensional?) spatial location between the particles.

Since the particles are charge-force carriers, Coulombs law will apply.

$F = k\frac{qQ}{r^2}$ which is analogous to Newtons gravitational force law $F = G\frac{mM}{r^2}$

By placing the third particle equidistant from the other two (equal +q charges), the forces acting on the third particle (-4q) will cancel leaving no net force and no acceleration. i.e. this is analogous to the Lagrange point in Newtonian gravitation.

The "evolution" of its (third particles) y-coordinate implies motion which can only occur if forces are acting to produce acceleration in the y-direction (unless there are also initial velocity conditions which are not previously stated).

The question also states 'non-linear differential which is saying the rate at which the y displacement changes w.r.t. change in x.

Unless the third particle is offset in the y-direction from the straight line path between the other two particles, we must conclude that NO net force acts on the centrally displaced particle and therefore there can be no acceleration in any direction.

As you can see, the question as originally stated is worded very ambiguously and, unless there is other information to clarify, any answers given will be a stab in the dark as we are trying to 'guess' the examiners intent.

Well, for a start it's a whole lot more information than the original question.

Start by deciphering the information given and working your way through to produce a model:

1) Construct the Cartesian 2-axis system. ( you could do it in 3-dimensions, but the question states cartesian x,y plane).

2) Note that the origin is also the equilibrium (zero potential) point between all three particles. i.e. the origin is centrally disposed with the positive charges equidistant either side.

3)The negative charge is free to move orthogonally to the x-axis and therefore can only be placed at one of two positions either side of the origin on the y-axis. (On a circumference if 3-axis system chosen).

4) The question wants you to apply the Newton's law analogy when working through the Coulomb solution (point charges). I take this to imply a Newtonian a mechanic solution is required and not anything more esoteric using non-inertial reference frames for instance. This will give the displacement from the origin (dy) in terms of the unit charge vector q assuming that charge is the standard multiple value e+. This calculation will also yield the magnitude of the y direction force vector. Also the charge mass is only stated as m and not actually given. So all of this implies they want a simple solution.

5)You will need to formulate an differential equation governing the displacement wrt (t) from the starting position of the -4p point charge. i.e. rate of change of distance wrt (t). Again using the Newton's law definition of force. (i.e. the equation will be a second order differential)

6) The solution will involve the negative charge motion in simple harmonic oscillation about the origin constrained to the y-axis. (2-dimensional solution). The quickest solution method here would be using the Laplace operator (no tedious integral and differential calculus required) to yield a solution of the form y = e^ax(Asinwt + Bcoswt).

7) Initial conditions for y(3t) substituted in above general solution.

So i have formed a differential equation, equating the electrostatic force acting downwards on the negative particle to the mass multiplied by the second order differential of displacement(y-direction) wrt (t).
Conditions are that at t=0, y(0)=0

Is that correct so far?

That sounds as if you are on the right track.

The electrostatic force acting on P3 should look something like:

(assuming point charges P1, P2 and P3 are represented by subscript 1, 2 & 3 respectively)

$\vec{F}_{e_y} = \vec{F}_{13} + \vec{F}_{23}$

$\vec{F}_{e_y} = \frac{1}{4 \pi \epsilon_{0}}(\frac{q_1 q_3}{(\sqrt{([d/2]^2+dy^2)^2}}\hat{r}_{13} + \frac{q_1 q_3}{(\sqrt{([d/2]^2+dy^2)^2}}\hat{r}_{23})$

and because the P3 -ve charge is symmetrically disposed between the two equal magnitude +ve charges, the e_x force vectors will cancel leaving

$\vec{F}_{e_y} = \vec{F}_{13}Cos\theta + \vec{F}_{23}Cos\theta$

where

$\theta = tan^{-1}(\frac{d/2}{dy}) = tan^{-1}(\frac{d}{2dy})$

(d=distance between +ve charges, dy=orthogonal displacement of the -ve charge from e_x along e_y)

i.e.

$\vec{F}_{e_y} = 2\vec{F}_{13}Cos\theta$ OR $2\vec{F}_{23}Cos\theta$

From Newton's laws we know that:

$F = ma$

$\vec{F}_{e_y} = m\frac{d^2y}{dt^2}$

$m\frac{d^2y}{dt^2} - \vec{F}_{e_y} = 0$

i.e.

this is in the form:

$\frac{d^2y}{dt^2} = \frac{-k}{m} x$ (constant of proportionality is $\frac{k}{m}$)

which is the same form as Hookes law

$F = -kx$


The general solution is a bit tedious so I will leave that to you (not least because using the tex editor is a pain for anything other than simple equations). You can do it in a number of ways depending on what you have already covered as part of your maths course.

Initial conditions are arbitrary because the actual solution is dependent on the initial displacement of P3 along the e_y axis and hence the maximum amplitude of the oscillation is dependent on this displacement since Acos(0) = A.

All solutions are sinusoidal (cosine) - max amplitude when passing through the e_y axis.

That sounds as if you are on the right track.

The electrostatic force acting on P3 should look something like:

(assuming point charges P1, P2 and P3 are represented by subscript 1, 2 & 3 respectively)

$\vec{F}_{e_y} = \vec{F}_{13} + \vec{F}_{23}$

$\vec{F}_{e_y} = \frac{1}{4 \pi \epsilon_{0}}(\frac{q_1 q_3}{(\sqrt{([d/2]^2+dy^2)^2}}\hat{r}_{13} + \frac{q_2 q_3}{(\sqrt{([d/2]^2+dy^2)^2}}\hat{r}_{23})$

and because the P3 -ve charge is symmetrically disposed between the two equal magnitude +ve charges, the e_x force vectors will cancel leaving

$\vec{F}_{e_y} = \vec{F}_{13}Cos\theta + \vec{F}_{23}Cos\theta$

where

$\theta = tan^{-1}(\frac{d/2}{dy}) = tan^{-1}(\frac{d}{2dy})$

(d=distance between +ve charges, dy=orthogonal displacement of the -ve charge from e_x along e_y)

i.e.

$\vec{F}_{e_y} = 2\vec{F}_{13}Cos\theta$ OR $2\vec{F}_{23}Cos\theta$

From Newton's laws we know that:

$F = ma$

$\vec{F}_{e_y} = m\frac{d^2y}{dt^2}$

$m\frac{d^2y}{dt^2} - \vec{F}_{e_y} = 0$

i.e.

this is in the form:

$\frac{d^2y}{dt^2} = \frac{-k}{m} x$ (constant of proportionality is $\frac{k}{m}$)

which is the same form as Hookes law

$F = -kx$


The general solution is a bit tedious so I will leave that to you (not least because using the tex editor is a pain for anything other than simple equations). You can do it in a number of ways depending on what you have already covered as part of your maths course.

Initial conditions are arbitrary because the actual solution is dependent on the initial displacement of P3 along the e_y axis and hence the maximum amplitude of the oscillation is dependent on this displacement since Acos(0) = A.

All solutions are sinusoidal (cosine) - max amplitude when passing through the e_y axis.