# Spring with charged objects

1. Jan 8, 2014

### Tanya Sharma

1. The problem statement, all variables and given/known data

Two equal and opposite charged objects of magnitude 2C and mass 2kg are attached to a spring of spring constant 4 N/m in the natural length of the spring. The system is suddenly placed in a uniform electric field of field strength 5 N/C in the direction of electric field such that the -ve charge is towards the origin of electric field. assuming the natural length of the spring to be large, find the maximum compression and elongation in the spring length.

2. Relevant equations

3. The attempt at a solution

First I tried to find the maximum extension.

Let charge -2 C be at a distance x1 and charge +2 be at a distance x2 from the origin .

Let the length of the spring be l .

The length of the spring at any instant is x2-x1 .

The extension in the spring would be x2-x1-l .

The force of attraction between the charges would be kq2/(x2-x1)2 i.e 16/(x2-x1)2

Force due to the external electric field will be of magnitude 10N .

EOM for mass 1 (i.e -2C ) will be k(x2-x1-l)-10+16/(x2-x1)2 = 2d2x1/dt2

EOM for mass 2 (i.e +2C ) will be 10-k(x2-x1-l)-16/(x2-x1)2 = 2d2x2/dt2

From the above we get ,

2(d2x2/dt2-d2x1/dt2)=20-2k(x2-x1-l)-32/(x2-x1)2

or , d2x2/dt2-d2x1/dt2=10-k(x2-x1-l)-16/(x2-x1)2

d2(x2-x1)/dt2=10-k(x2-x1-l)-16/(x2-x1)2

Now putting x2-x1 = z ,

We have d2z/dt2=10-k(z-l)-16/z2 .Now I need to maximize z .

Is it the correct way to approach the problem ?

I would be grateful if someone could help me with the problem .

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2. Jan 8, 2014

### Saitama

Hello Tanya!

I am not sure but I think you are required to use the assumption that the natural length of spring is large to cancel one of the factors.

Also, how do you get kq^2=20?

3. Jan 8, 2014

### voko

For this kind of problems (two interacting bodies) it is usually simplest to convert to the CoM frame and use conservation of total momentum in the CoM frame. This separates coordinates into the coordinates of the CoM and the distance between the bodies.

4. Jan 8, 2014

### Tanya Sharma

Does that mean the force between the charges doesn't come in picture ? Do I have to neglect this force ?

I have fixed the error .kq2=16

Have I approached it correctly ?

5. Jan 8, 2014

### Saitama

This is what I think but as I said, I am not sure. :)

Still incorrect. It looks to me as if you are confusing the Coulomb's constant with the spring constant.

6. Jan 8, 2014

### Tanya Sharma

. Forgive me for that .What next ?

7. Jan 8, 2014

### Saitama

You can use the substitution $x_2-x_1-l=z$ but your substitution is fine too.

What do you get if you solve the differential equation? I suggest to denote everything by symbols and substitute the values later.

Btw, voko's method is a nice one, you can use energy conservation in the CM frame, no need of differential equations. :)

8. Jan 9, 2014

### consciousness

Work Energy theorem is the best approach according to me.
Since the magnitude of charge and the masses are same, these objects will move symmetrically in opposite directions. Therefore, when maximum compression and elongation occur, both the objects will be at rest. So you just have to consider
(i)Spring potential energy
(ii)Interaction potential energy
(iii)Work done by external force
while making the equations.

9. Jan 9, 2014

### Tanya Sharma

I understand work energy approach is better .But since I started off with DE's ,want to finish it first.

The DE is

d2(x2-x1)/dt2=10-k(x2-x1-l)

Now putting x2-x1 = z ,

We have d2z/dt2=10-k(z-l)

or , d2z/dt2+kz-(kl+10) = 0

The solution is z = C1cos(√kt) + C2sin(√kt) + (l + 10/k)

Does that make sense ?

10. Jan 9, 2014

### voko

"assuming the natural length of the spring to be large" means that you are supposed to approach the problem assuming that $s = z - l << l$. It does not mean that the term for the electrical interaction between the charges is completely neglected. It means that it must be approximated with a term linear in $s$.

11. Jan 9, 2014

### Tanya Sharma

Hi voko...

Suppose we neglect the force between the charges , does post #9 makes sense ?

12. Jan 9, 2014

### voko

In the sense whether the solution of the equation in #9 is correct, yes.

13. Jan 9, 2014

### Tanya Sharma

Okay...

So ,z = C1cos(√kt) + C2sin(√kt) + (l + 10/k)

Now ,using z(0) = l we get C1 = -10/k .

How do I find C2 ?

14. Jan 9, 2014

### voko

What about $\dot z(0)$?

15. Jan 9, 2014

### Tanya Sharma

That gives C2 = 0 .

So,z =(-10/k)cos(√kt) + (l+10/k) . Is it correct ?

16. Jan 9, 2014

### voko

The equation from #9, which you have solved, is not a correct equation for the original problem. Yet you used the initial conditions from the original problem to determine the coefficients in its solution. That does not look correct to me.

17. Jan 9, 2014

### Saitama

Hi voko!

This is the DE we have:
$$-2k(x_2-x_1-l)-\frac{2Kq^2}{(x_2-x_1)^2}+2qE=m\left(\frac{d^2x_2}{dt^2}-\frac{d^2x_1}{dt^2}\right)$$

If we go with Tanya's substitution $x_2-x_1=z$, I don't see how we get z-l in the denominator of electric force.

18. Jan 9, 2014

### voko

I suggest that you first rewrite that equation in terms of $s$. Then think how you can simplify the equation if $s \ll l$.

19. Jan 9, 2014

### Saitama

OK, so I switch to the substitution $x_2-x_1-l=s$, then I have $\ddot{x_2}-\ddot{x_1}=\ddot{s}$, hence
$$-2ks-\frac{2Kq^2}{(s+l)^2}+2qE=m\frac{d^2s}{dt^2}$$
$$\Rightarrow -2ks-\frac{2Kq^2}{l^2}\left(1+\frac{s}{l}\right)^{-2}+2qE=m\frac{d^2s}{dt^2}$$
$$\Rightarrow m\frac{d^2s}{dt^2}=-2ks+\frac{4Kq^2s}{l^3}+2qE-\frac{2Kq^2}{l^2}$$
Do I have to solve the above DE?

20. Jan 9, 2014

### voko

The problem does not want you to solve differential equations. It wants you to find the minimum and maximum elongations.

Note that the equation you obtained can be re-interpreted as an equation for a massive particle in a force field, where the force depends linearly on the position. That force is potential, so you could find the corresponding potential energy. Then you can write down the equation for conservation of energy, and determine the min/max elongation from it.