Time of approach of two oppositely charged particle

[Satvik Pandey]

Homework Statement

Two point charges q1=1\muC and q2=-1\muC with mass 1g each are at distance 1m from each other. The charges are released from rest at . Find the time t in seconds at which they collide.

Homework Equations

F=kQ1*Q2/r^2.

The Attempt at a Solution

Let the dist. of separation be r and the mass of charged be m.
I think they will meet at the middle of separation.Force on q1=kq^2/r^2 so the acceleration =kq^2/r^2*m.But in this case acceleration is not constant.So I have to use calculus here.
Let the charge q1 be at origin and at a distance x from origin acceleration=kq^2/(r-2x)^2*m( 'r-2x' because if q1 will move x meter towards q2 then q2 will also move x metre towards q1 so the length of separation will decrease by 2x) .
For a small displacement dx this acceleration is constant.I tried to use S=ut+at^2/2.But I am unable to find u.Please help.

 

[ehild]

Have you studied differential equations?

When setting up the solution, draw a picture first and show how you define the variables in that picture.

The particles move symmetrically with respect to the centre. If one is at x m distance from the centre, how far is the other? And what is the distance between them?

ehild

 

[Satvik Pandey]

The particles move symmetrically with respect to the centre. If one is at x m distance from the centre, how far is the other? And what is the distance between them?

ehild

The other should also be x m distance from the center.So the distance between them should be 2x.

 

[Satvik Pandey]

I should have choose origin at the center.Now at a distance x from the origin acceleration of q1 will be kq^2/(4x^2*m).Here m is for mass of particle.
I have made this figure

.

 

[ehild]

I should have choose origin at the center.Now at a distance x from the origin acceleration of q1 will be kq^2/(4x^2*m).Here m is for mass of particle.
I have made this figure View attachment 71218.

Good. How the acceleration is related to the second derivative of x?

ehild

 

[Satvik Pandey]

Good. How the acceleration is related to the second derivative of x?

ehild

\frac{d^{2}x}{dt^{2}}=acceleration

 

[ehild]

Write the equation of motion ma=F in terms of x. You wrote that the acceleration is kq^2/(4mx²). It is positive. That means x increasing with time. Is it true? Write the equation of motion in terms of x.

ehild

 

[Satvik Pandey]

Write the equation of motion ma=F in terms of x. You wrote that the acceleration is kq^2/(4mx²). It is positive. That means x increasing with time. Is it true? Write the equation of motion in terms of x.

ehild

kq^{2}/4x^{2}=m*d^2x/dt^2
I think x decreases with time because as the time passes both the particle move toward origin so their distance of separation decreases.

 

[ehild]

Look at m1. The force acts to the right. The acceleration is to the right. Does x increase?

ehild

 

[Satvik Pandey]

According to figure which I have made in post#4 it seems that x decreases and acceleration increases because x is measured from origin and in this figure origin is at the middle of the line of separation.

 

[ehild]

Yes. So -kq2/4x2=m*d^2x/dt^2
That is a differential equation, it is not easy to solve. You can go one step ahead by applying Conservation of Energy. Then you get the speed as function of x. Try.
ehild

 

[Satvik Pandey]

Yes. So -kq2/4x2=m*d^2x/dt^2
That is a differential equation, it is not easy to solve. You can go one step ahead by applying Conservation of Energy. Then you get the speed as function of x. Try.
ehild

I have not applied conservation of energy concept to system of particles earlier so this is my first time.
Initially E(potential)=-\frac{kq^{2}}{r} (I got -ve sign as the charge at the left is negative)
Now at x kinetic energy of system = kinetic energy of the CM+ kinetic energy of all particles relative to the CM.
E_{k} at x=mv^{2}
E_{p} at x = -\frac{kq^{2}}{2x}
so
-\frac{kq^{2}}{r}=mv^{2}+ -\frac{kq^{2}}{2x}.

v=sq.root of \frac{kq^{2}(r-2x)}{2mxr}.
the figure is here View attachment 71218.V is the velocities of particles at x.

 

[Tanya Sharma]

Now write v =dx/dt and integrate to find x as function of time .

 

[ehild]

Be careful with the signs. x is the distance between the charges. It decreases with time, dx/dt<0. v is the speed of both particles, a positive quantity.

ehild

 

[Satvik Pandey]

Well I got
-∫\frac{√2x}{√(1-2x)}dx=∫3*10^6 dt.
I have some problem in solving LHS.I substituted 2x=a^2.
And I got dx=ada.On putting values I got
∫\frac{a^2}{√(1-a^2}da.
Don't know what to do next.

 

[Tanya Sharma]

This will not work . Try some trigonometric substitution and see which one fits in.

 

[Satvik Pandey]

This will not work . Try some trigonometric substitution and see which one fits in.

If denominator is in the √a^2-x^2 and numerator is dx then we substitute x as asinθ.I don't have very much knowledge of calculus.I know only some basics.Will the substitution work.

 

[Tanya Sharma]

While performing integration using substitution method you need to substitute values of both x as well as dx in terms of θ.

For example if you are using the substitution x= asinθ , then dx = (acosθ)dθ .

 

[ehild]

Well I got
-∫\frac{√2x}{√(1-2x)}dx=∫3*10^6 dt.

You can cheat a bit... It is not an easy integral. Use wolframalpha.com

Check the right-hand side, it does not look correct.

Spoiler

http://www.wolframalpha.com/input/?i=int_0.5^0%28sqrt%282x%2F%281-2x%29%29dx%29

ehild

 

[Tanya Sharma]

Yes . The integral is not easy .

 

[Satvik Pandey]

While performing integration using substitution method you need to substitute values of both x as well as dx in terms of θ.

For example if you are using the substitution x= asinθ , then dx = (acosθ)dθ .

Well I substituted 2x=sin^{2}θ.
so dx=sinθcosθdθ.
On substituting the values
∫ √(sin^2θ)/√(1-sin^2θ) dθ

=∫sin^2θ dθ

=θ/2 - sin2θ/4

 

[ehild]

Very nice! Now the limits. What are the limits of integration?

ehild

 

[Satvik Pandey]

Check the right-hand side, it does not look correct.


ehild

\frac{dx}{dt}=√\frac{kq^{2}(r-2x)}{2mxr}

On putting the values I got RHS as

√9*10^9*1(1-2x)/2*10^{-3}*x*1
( the mass was given in grams and that has to converted into kg)
so dx/dt=√( 9* 10^12(1-2x)/2x)

so I got this ∫√2x/√(1−2x)dx=∫3*10^6 dt.

 

[ehild]

I meant the constant sqrt(kq^2/(mr)). You seem to omit q^2. q=1 μC.

ehild

 

[Satvik Pandey]

Very nice! Now the limits. What are the limits of integration?

ehild

Earlier the limits of x are from o.5 to 0 but 2x=sin^2 θ
so now the limits are from 90 degrees to 0

 

[ehild]

Yes, but you should use radians.

ehild

 

[Satvik Pandey]

I meant the constant sqrt(kq^2/(mr)).

ehild

Was the equation in post#12 not correct?

 

[Satvik Pandey]

Yes, but you should use radians.

ehild

So the limits are from 'pi'/2 to 0

 

[ehild]

Was the equation in post#12 not correct?

I was correct but that in post #15 was wrong. The number in the right-hand side.

ehild

 

[ehild]

So the limits are from 'pi'/2 to 0

Excellent!

ehild

 

[Satvik Pandey]

I meant the constant sqrt(kq^2/(mr)). You seem to omit q^2. q=1 μC.

ehild

Yes I forgot that q=1 μC.
It should be ∫√2x/√(1−2x)dx=∫3 dt.

 

[ehild]

Yes. So what is t then?

ehild

 

[Satvik Pandey]

Yes. So what is t then?

ehild

LHS is 0.785398 and RHS is 3t

So I got t=0.2619sec but the answer is 0.74second.
Is the answer correct?

 

[ehild]

I think your solution is correct.

ehild

 

[Tanya Sharma]

I am also getting $\frac{\pi}{12} sec$ as the answer .

 

[Satvik Pandey]

There is a hint given below with the question.
It is ----"Do it without integrating using Kepler's Law."
But Kepler's law are applicable to the planet moving in elliptical orbits.
Can it be applied here?

 

[ehild]

Yes, it can be. The electric field is also a central force, like gravity.
The two-body problem can be reduced by a single-body problem with the reduced mass method. The problem is equivalent if we consider a virtual body of reduced mass (μ=1/(1/m1 + 1/m2) about the centre of the force, where the distance from the centre is equal to the distance between the bodies in the original problem.
The straight line can be considered as a very elongated ellipse. At the limit, the focal point is at the opposite end of the distorted ellipse.
Kepler's third law states that the time period depends only on the average distance from the central body, on the semi-mayor axis.
You can determine what would be the period of a circular orbit with the radius as the semi-mayor axis.

ehild

 

[rcgldr]

An old thread for objects accelerating towards each other due to gravity, showing Kepler method (which still used integration) and method based on integrating acceleration to get velocity (rate of closure), then integrating velocity to get position.

https://www.physicsforums.com/showthread.php?t=635188

 

[Satvik Pandey]

Just before the collision their acceleration is infinite.Are their velocities also infinite just before the collision?

 

[Satvik Pandey]

If the charges on the particles would have been different then there would not be much problem I think.
But what would happen if their masses were different.I think in this, they will not meet at the center of path.I think they will meet at the CM of the system.But the motion of the particles will not be symmetrical in this case.How this will be solved?

 

[ehild]

Just before the collision their acceleration is infinite.Are their velocities also infinite just before the collision?

Yes. The total energy is finite, the potential energy is negative infinite at collision so the E must be positive infinite.

ehild

 

[ehild]

You calculated with the distance of the particles previously. You can work with that distance again.

Chosse a system of coordinates with the CM as origin. Let be x₁ and x₂ the coordinates of the particles. Then m₁x₁+m₂x₂=0 x₁=-x₂m₂/(m₁)

Choose a new variable, r=x₂-x₁. Then x₁=-rm₂/(m₁+m₂) and x₂=rm₁/(m₁+m₂)
The velocities are
v₁=-dr/dt m₂/(m₁+m₂)
v₂= dr/dt m₁/(m₁+m₂)

The kinetic energy is KE=0.5 μ (dr/dt)² where μ=m₁m₂/(m₁+m₂). μ is called the reduced mass of the particles.

The potential energy is PE=-q1q2/r. Initially the distance is R and the particles are in rest.

Write up the equation for conservation of energy: 0.5 μ (dr/dt)² - q1q2/r=-q1q2/R and solve with the same method you have applied for the OP.

ehild

 

[Tanya Sharma]

The problem is equivalent if we consider a virtual body of reduced mass (μ=1/(1/m1 + 1/m2) about the centre of the force, where the distance from the centre is equal to the distance between the bodies in the original problem.

There are three terms involved Center of Mass , Center of force and Central body (M+m) . Isn't the Central body also the Center of force ? What are the roles of Center of Mass and Central body in the reduced mass approach ?

The straight line can be considered as a very elongated ellipse. At the limit, the focal point is at the opposite end of the distorted ellipse.

Could you explain it a bit more ? I am not very clear with this idea of masses orbiting in straight line .

Thanks

 

[Satvik Pandey]

Yes. The total energy is finite, the potential energy is negative infinite at collision so the E must be positive infinite.

ehild

As the particles starts coming towards each other their potential begins to increase(E_{p}=kq^2/r so if r decreases E_{p} increases) and their kinetic energy also increases (because acceleration increases hence velocity also increases).Doesn't it violates law the of conservation of energy?

 

[Satvik Pandey]

Yes, it can be. The electric field is also a central force, like gravity.
The two-body problem can be reduced by a single-body problem with the reduced mass method. The problem is equivalent if we consider a virtual body of reduced mass (μ=1/(1/m1 + 1/m2) about the centre of the force where the distance from the centre is equal to the distance between the bodies in the original problem
ehild

What is centre of force?

where the distance from the centre is equal to the distance between the bodies in the original problem.
ehild

I don't understand what you trying to say from this line.

 

[ehild]

There are three terms involved Center of Mass , Center of force and Central body (M+m) . Isn't the Central body also the Center of force ? What are the roles of Center of Mass and Central body in the reduced mass approach ?

In case when the force of interaction is the Coulomb force, there is no "central mass".

Could you explain it a bit more ? I am not very clear with this idea of masses orbiting in straight line .

Thanks

A straight line is just a very - very elongated ellipse. At the limit, it becomes a section of a straight line. The virtual body is at one end of the section initially, and the focus is on the other end.

ehild

 

[ehild]

As the particles starts coming towards each other their potential begins to increase(E_{p}=kq^2/r so if r decreases E_{p} increases) and their kinetic energy also increases (because acceleration increases hence velocity also increases).Doesn't it violates law the of conservation of energy?

The potential energy is negative. Ep = - kq²/r . It decreases if r decreases.

ehild

 

[ehild]

What is centre of force?


I don't understand what you trying to say from this line.

Both the Coulomb force and gravity are central forces. I used the term "centre of force" in the frames of the reduced mass method. It relates the original problem of a single particle orbiting around a centre. The force points towards that centre and its magnitude is given, kq²/r² this time. The distance from the centre of that virtual particle is the distance between the real particles.

Browse "two-body problem" and "reduced mass".

ehild

 

[Satvik Pandey]

I searched and found that central forces depends on the distance between the particles and they are directed along the line joining them.
I found reduce mass of two particles system is m1*m2/m1+m2 and this is an imaginary particle.But what is the position of this imaginary particle and how this will help me in solving problem which I have posted in post#40.

 

[Tanya Sharma]

But what is the position of this imaginary particle and how this will help me in solving problem which I have posted in post#40.

The reduced mass would be at the same distance as between the two charged particles in the original problem.Initially it is at a distance of 1 m .

ehild has given you pretty much everything in post#42 to solve the problem.

Write up the equation for conservation of energy: 0.5 μ (dr/dt)² - q1q2/r=-q1q2/R and solve with the same method you have applied for the OP.

Here 'r' is the instantaneous separation .It will be solved on similar lines to what you have done earlier.