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