Electrostatics Problem -Two moving charged particles

[Tanya Sharma]

Two small particles have electric charges of equal magnitude and opposite signs. The masses of the particles are m and 2m. Initially, the distance between the particles is d, and the velocities of the particles have equal magnitude v. However, the velocity of particle 2m is directed away from particle m, whereas the velocity of particle m is directed perpendicular to the line connecting the particles. In the subsequent motion of the particles, they are found to be at a distance 3d from each other—twice. Find the possible values of the charge of each particle.

Homework Equations

The Attempt at a Solution

My approach is to find the maximum distance ‘x ‘ between the charges and impose the condition that x>d ,as then the charges will be at a distance d- twice .

I will work in CM frame of reference.
$V_{cm} = \frac{1}{3m}(2mv\hat{i}+mv\hat{j})$

$V_{cm} = \frac{2v}{3}\hat{i}+\frac{v}{3}\hat{j}$

Now , Let $V_{2i}$ be initial velocity of 2m and $V_{1i}$ be initial velocity of m in CM frame.

$V_{2i} = \frac{v}{3}\hat{i}-\frac{v}{3}\hat{j}$

$V_{i} = \frac{-2v}{3}\hat{i}+\frac{2v}{3}\hat{j}$

Let $V_{2f}$ be initial velocity of 2m and $V_{1f}$ be velocity of m in CM frame ,when the particles are at maximum distance.

Since no external forces are acting,linear momentum is conserved .

Initial momentum in CM frame = Final momentum in CM frame = 0

$2mV_{2f} = mV_{1f}$

Let $V_{1f} = v'$ ,so $V_{1f} = \frac{v'}{2}$

Since no external torques are present ,angular momentum is conserved in CM frame .

$2m\frac{v}{3}\frac{d}{3} + m\frac{2v}{3}\frac{2d}{3} = (2m)\frac{v'}{2}\frac{x}{3} + (m)(v')\frac{2x}{3}$

This gives $v' = \frac{2vd}{3x}$

Now applying conservation of energy ,

$\frac{1}{2}m(\frac{2v}{3})^2+\frac{1}{2}2m(\frac{v}{3})^2 - \frac{kq^2}{d} = \frac{1}{2}mv'^2+\frac{1}{2}2m(\frac{v'}{2})^2 - \frac{kq^2}{x}$

$\frac{1}{2}m(\frac{2v}{3})^2+\frac{1}{2}2m(\frac{v}{3})^2 - \frac{kq^2}{d} = \frac{1}{2}m(\frac{2vd}{3x})^2+\frac{1}{2}2m(\frac{1}{2}\frac{2vd}{3x})^2 - \frac{kq^2}{x}$

$\frac{1}{3}mv^2- \frac{kq^2}{d} = \frac{1}{3}\frac{mv^2d^2}{x^2}- \frac{kq^2}{x}$

$\frac{mv^2d-3kq^2}{3d} = \frac{mv^2d^2-3kq^2x}{3x^2}$

I am not sure whether I am approaching the problem correctly .I would be grateful if somebody could help me with the problem.

 

[haruspex]

My approach is to find the maximum distance ‘x ‘ between the charges and impose the condition that x>d ,as then the charges will be at a distance d- twice .

Good approach, but you mean 3d - twice.
There is another constraint on x, though.

$2mV_{2f} = mV_{1f}$

Don't you mean $2mV_{2f} = -mV_{1f}$

Let $V_{1f} = v'$ ,so $V_{1f} = \frac{v'}{2}$

This is confusing. v is a scalar, but v' is a vector?

Since no external torques are present ,angular momentum is conserved in CM frame .

$2m\frac{v}{3}\frac{d}{3} + m\frac{2v}{3}\frac{2d}{3} = (2m)\frac{v'}{2}\frac{x}{3} + (m)(v')\frac{2x}{3}$

Not sure how you are deriving these expressions for angular momentum. In the initial condition, 2m has no moment about CM.

 

[Tanya Sharma]

Good approach, but you mean 3d - twice.

Yes ,I meant 3d twice .

There is another constraint on x, though.

I do not know .

Don't you mean $2mV_{2f} = -mV_{1f}$

This is confusing. v is a scalar, but v' is a vector?

I was equating the speeds of the two particles at maximum separation.

Not sure how you are deriving these expressions for angular momentum. In the initial condition, 2m has no moment about CM.

$V_{2i} = \frac{v}{3}\hat{i}-\frac{v}{3}\hat{j}$

Here $V_{2i}$ is the initial velocity of 2m in CM frame .

$\vec{r} = \frac{d}{3}\hat{i}$

By using the definition of angular momentum $m\vec{r} \times {\vec{V}_{2i}}$ - Why would 2m not have angular momentum about the CM ?

 

[TSny]

Since no external forces are acting,linear momentum is conserved .

Initial momentum in CM frame = Final momentum in CM frame = 0

Let $V_{1f} = v'$ ,so $V_{1f} = \frac{v'}{2}$

Typo: V_2f = v'
[EDIT: I should have written V_2f = v'/2 as you noted in post #5. (Typo of a typo )]

Now applying conservation of energy ,

$\frac{1}{2}m(\frac{2v}{3})^2+\frac{1}{2}2m(\frac{v}{3})^2 - \frac{kq^2}{d} = \frac{1}{2}mv'^2+\frac{1}{2}2m(\frac{v'}{2})^2 - \frac{kq^2}{x}$

Did you use the correct values for the initial speeds?

Otherwise, I think your work is OK. What do you get for the charge q if x is set equal to the critical value 3d?

 

[Tanya Sharma]

Typo: V_2f = v'

I think it should be $V_{1f} = v'$ and $V_{2f} = \frac{v'}{2}$

Did you use the correct values for the initial speeds?

It should be

$$ \frac{1}{2}m(\frac{2\sqrt{2}v}{3})^2+\frac{1}{2}2m(\frac{\sqrt{2}v}{3})^2 - \frac{kq^2}{d} = \frac{1}{2}mv'^2+\frac{1}{2}2m(\frac{v'}{2})^2 - \frac{kq^2}{x} $$

 

[Tanya Sharma]

What do you get for the charge q if x is set equal to the critical value 3d?

$$ q = \sqrt{\frac{17mv^2d}{18k}} $$

Edit:Removed error

 

[TSny]

$$ q = \sqrt{\frac{17mv^2d}{18kd}} $$

I think that's right except you have too many d's in your answer.

 

[Tanya Sharma]

OK.

But this gives me the upper bound of the value of charge .How should I get the lower bound ?

Another thing , how do I know whether the result just obtained is lower or upper bound ?

 

[TSny]

OK.

But this gives me the upper bound of the value of charge .How should I get the lower bound ?

Another thing , how do I know whether the result just obtained is lower or upper bound ?

You should be able to answer these questions if you think about the different types of motion that can occur for an inverse-square force of attraction.

 

[haruspex]

How should I get the lower bound ?

As I wrote in post #2, there is another constraint on x. How large can x be? How large can any number be? Think big, really big!

 

[Tanya Sharma]

As I wrote in post #2, there is another constraint on x.

I haven't been able to think about it.

How large can x be? How large can any number be? Think big, really big!

Are you indicating to the symbol in your "Thanks" badge .

 

[ehild]

You should be able to answer these questions if you think about the different types of motion that can occur for an inverse-square force of attraction.

Adding to TSny's hint, this two-body problem can be reduced to a one-body problem with the reduced mass approach, you are familiar with. You get the same equation for the difference of the position vectors as in case of the Kepler problem or the Hydrogen atom. A particle in an attracting, central, inverse-square force field. There are two conserving quantities, one is energy, what is the other one?
What kind of orbits are possible? What is the condition that the orbit is bounded? When can the particle move infinitely far from the centre?

It is not quite clear form the text of the problem, that the particles are really only twice at distance 3d, or they are twice at that distance along their repeating orbits.

ehild

 

[haruspex]

Are you indicating to the symbol in your "Thanks" badge .

That's the one. What inequality relates x to that?

 

[Tanya Sharma]

As I wrote in post #2, there is another constraint on x. How large can x be? How large can any number be? Think big, really big!

That's the one. What inequality relates x to that?

x<∞ ,but I don't know the value of x ?

You should be able to answer these questions if you think about the different types of motion that can occur for an inverse-square force of attraction.

Circular and Elliptical ,just as in planetary motion .

A particle in an attracting, central, inverse-square force field. There are two conserving quantities, one is energy, what is the other one?

Angular momentum .

What kind of orbits are possible? What is the condition that the orbit is bounded? When can the particle move infinitely far from the centre?

Circular and elliptical orbits are possible .The energy should be negative.When the particle has non negative energy .

It is not quite clear form the text of the problem, that the particles are really only twice at distance 3d, or they are twice at that distance along their repeating orbits.

ehild

If the two particles collide with each other then it should be former .If they are orbiting around the common CM then it should be latter.

 

[Tanya Sharma]

If I put the initial energy as negative I get the lower bound i.e E1 <0

$$\frac{1}{2}m(\frac{2\sqrt{2}v}{3})^2+\frac{1}{2}2m(\frac{\sqrt{2}v}{3}) ^2 - \frac{kq^2}{d} < 0 $$

This gives $$q > \sqrt{\frac{2mv^2d}{3k}}$$

Does this make sense ?

 

[ehild]

x<∞ ,but I don't know the value of x ?

It is less than infinity

Circular and Elliptical ,just as in planetary motion .

Yes, the planets are on circular and elliptical orbits, but what other orbits are possible?

Angular momentum .

yes, the angular momentum is also conserved and it is not zero.

Circular and elliptical orbits are possible .The energy should be negative.When the particle has non negative energy .

And what happens if the energy is zero or positive?

If the two particles collide with each other then it should be former .If they are orbiting around the common CM then it should be latter.

Can the particles collide if the angular momentum of the system with respect to the CM differs from zero? What would it be in the instant of collision?

ehild

 

[Tanya Sharma]

Please have a look at post#15

 

[ehild]

Explain why the energy must be negative? ehild

 

[Tanya Sharma]

So that the two particles are under the influence of each others electrostatic force i.e are bound with each other just like Earth and sun .

 

[ehild]

What about a spaceship that leaves the solar system? Or a not-returning comet? When the attracting force is small and the speeds are great?

ehild

 

[Tanya Sharma]

The total energy of the system becomes non negative at a certain point in space and the spaceship no longer orbits i.e it is out of the gravitational field .

 

[haruspex]

If I put the initial energy as negative I get the lower bound i.e E1 <0

$$\frac{1}{2}m(\frac{2\sqrt{2}v}{3})^2+\frac{1}{2}2m(\frac{\sqrt{2}v}{3}) ^2 - \frac{kq^2}{d} < 0 $$

This gives $$q > \sqrt{\frac{2mv^2d}{3k}}$$

Does this make sense ?

Yes, but see below.

ehild: Tanya is saying that if we take the PE as zero at infinity, then failure to reach infinity means the initial total energy is less than zero. Of course, it isn't quite that simple because it overlooks that there could still be some tangential KE that is not converted to PE. However, going back to the equations and letting x tend to infinity from below reveals that this does not change the answer.

 

[ehild]

Yes, but the spaceship burns all its fuel still in the Solar System, and its energy is conserved after that.

So if the energy is non-negative the orbit extends to infinity: parabolic or hyperbolic. Can it happen in this set-up?

ehild

 

[ehild]

Yes, but see below.

ehild: Tanya is saying that if we take the PE as zero at infinity, then failure to reach infinity means the initial total energy is less than zero. Of course, it isn't quite that simple because it overlooks that there could still be some tangential KE that is not converted to PE. However, going back to the equations and letting x tend to infinity from below reveals that this does not change the answer.

I asked Tania to explain why the particles can not be separated to infinite distance. You get bounded orbits if the total energy is negative. Otherwise the orbit is parabola or hyperbole.

What is x? The distance between the particles?

ehild

 

[haruspex]

What is x? The distance between the particles?

In the OP it is defined as the max distance between them.

 

[ehild]

In the OP it is defined as the max distance between them.

No, 3d is not the maximum distance.

In the subsequent motion of the particles, they are found to be at a distance 3d from each other—twice. Find the possible values of the charge of each particle.

ehild

 

[haruspex]

No, 3d is not the maximum distance.

I didn't say 3d was the max distance. I said x is defined in the OP as the max distance. An equation is found for x, and the inequalities 3d < x < ∞ are applied.

 

[ehild]

At last I have found in the OP what x meant.

I missed a clear "discussion" from Tania's solution. What is the reason to assume negative energy?
The solution is correct otherwise, but the students need to explain the steps they apply.

And I would prefer the reduced mass method.

Instead of handling the particles separately, with their own position vectors and velocities, use those of the CM and the relative position of the particles $\vec r = \vec r_2 - \vec r_1$

The CM moves with constant velocity, and $\vec r$ obeys the equation

$μ \ddot {\vec r }= -k\frac {Q^2}{r^3}\vec r$

where μ is the reduced mass and r is the distance between the particles. In this case, $μ = 2/3 m$. It is a simple Kepler problem with the initial conditions

$r_i= d \vec i$ and $\vec v_i= v(\vec i - \vec j)$.

From the initial data, the energy is $E=μv^2-k\frac{q^2}{d}$ and the angular momentum is $L=-μdv$.

We know that the possible orbits are ellipse(circle), parabola, and hyperbole . The problem said that the distance between the particles is 3d twice, for the imaginary particle, it means distance from the focal point. In case of unbounded orbits, there is only one extremal distance, a minimum. If the distance increased from d to 3d, it never decreases again, the particle goes to infinity.
So the particle moves along an elliptic orbit. Energy is conserved and angular momentum is conserved. The energy is negative:

$E=μv^2-k\frac{q^2}{d}<0 \rightarrow q^2>\frac{μv^2d}{k}$.

The other bound for q is obtained from the condition that the particle goes farther than 3d.

Working with polar coordinates

$\vec v= \dot r \hat r + r\dot θ \hat θ$ , so

$E=0.5 μ \left( \dot r ^2+(r\dot θ )^2\right) -k\frac{q^2}{r}$

and the angular momentum is

$L= μ r^2 \dot θ \rightarrow \dot θ=\frac{L}{μr^2}=\frac{-dv}{r^2}$

Substituting for $\ddot θ$ into the formula for energy:

$E=0.5 μ \left ( \dot r ^2+\left(\frac{-dv}{r} )^2\right) \right)-k\frac{q^2}{r}$.

Applying conservation of energy and substituting r=3d,

$E=0.5 μ \left ( \dot r ^2+\frac{v^2}{9} \right)-k\frac{q^2}{3d}=μv^2-k\frac{q^2}{d}$

3d is not the maximum distance, so the other bound for q² is obtained by using that $(\dot r)^2>0$.

ehild