Finding the distance traveled by a charged object

[afk94]

There are two electrically charged objects that are placed 6 cm away from each other. Each object has an electrical charge of +1 microColoumbs and a mass of 1 milligram. If one object is fixed in place, how far does the other go in 2 milliseconds?

So I'm given the charge, distance between the charges, the mass of the charges, and the change in time.
1. I started by finding the Force of each charge on the other using F=k qA*qB/d^2. I got 2.5 N as my answer.
2. I then used F=ma or more specifically a=F/m to find the acceleration which I got as 2.5*10^6 N/kg.
3. I then solved for final Velocity by using Vf = Vi +at (I assumed Vi was 0) and got 5*10^3 m/s
4. I then solved for Average velocity by using Vavg = 1/2 (Vi + Vf) and got 2.5*10^3 m/s
5. I then found the distance by using d = Vavg *Δt and got 5 m.

Can anyone tell me if I got the correct answer and if I used the correct steps. If not, what did I do wrong? Thanks

 

[jackarms]

Careful, the equations you're using are for constant acceleration, and that's not the case here. As one charge moves away from the other, the distance between them changes, so the force does as well, and in turn the acceleration. That does make the problem more difficult -- looks like you'll have to use integration from the force law.

 

[afk94]

Careful, the equations you're using are for constant acceleration, and that's not the case here. As one charge moves away from the other, the distance between them changes, so the force does as well, and in turn the acceleration. That does make the problem more difficult -- looks like you'll have to use integration from the force law.

Is there an equation for acceleration that isn't constant?

 

[jackarms]

Yes, you can get it from the force and the mass:

$$a = \frac{F}{m} = \frac{q_{1}q_{2}}{4\pi \epsilon_{0}d^{2}m_{1}}$$,

where [itex]m_{1}[/itex] is the one moving, and [itex]d[/itex] is the distance between them.

 

[afk94]

What does ϵ0 represent?

 

[jackarms]

Maybe you haven't it seen it before: it just a constant that comes from Coulomb's Law:

$$F = \frac{1}{4\pi \epsilon_{0}} \cdot \frac{q_{1}q_{2}}{r^{2}}$$

EDIT: I fixed the acceleration equation above too -- had masses on top instead of charges.

 

[afk94]

Maybe you haven't it seen it before: it just a constant that comes from Coulomb's Law:

$$F = \frac{1}{4\pi \epsilon_{0}} \cdot \frac{q_{1}q_{2}}{r^{2}}$$

EDIT: I fixed the acceleration equation above too -- had masses on top instead of charges.

Ahh ok so is 1/4∏ε0 the same as k? In my physics class, the professor always gave us k as a constant of 9*10^9, but never told us where that came from.

 

[jackarms]

That's right -- k is just written in terms of another constant, [itex]\epsilon_{0}[/itex], which is called permitivity, if you're interested. My teacher uses both forms, since [itex]\epsilon_{0}[/itex] shows up in other equations as well.

 

[afk94]

So the formula for a changing acceleration would be a= q1*q2/ k*d^2* m1?
If I solve for acceleration using that formula, would I have to use different formulas to find the velocities and distance traveled or can I use the same ones I originally used?
I appreciate your help

 

[jackarms]

You have to use different formulas, since the acceleration changes. It ends up being a differential equation, since the acceleration (second derivative of position) is proportional to the inverse of position squared. Have you done this type of problem before?

 

[afk94]

You have to use different formulas, since the acceleration changes. It ends up being a differential equation, since the acceleration (second derivative of position) is proportional to the inverse of position squared. Have you done this type of problem before?

Unfortunately I have not. Everything I've done so far has involved standard kinematic equations and assuming acceleration is constant. I've never done anything where acceleration isn't constant.

 

[jackarms]

Yeah, this is a type of Kepler problem, and one that isn't easily solved. I can't imagine you would be expected to solve something like this. Where does this problem come from, might I ask?

 

[afk94]

My professor gave it to my class as a "project." When I asked him whether or not we should assume acceleration to be constant he said "that is something for you to decide." Which I'm assuming means yes because if we don't assume acceleration to be constant then there's no way that anyone in the class would have the knowledge or skills to do a Kepler problem, but who knows.

 

[afk94]

Yeah, this is a type of Kepler problem, and one that isn't easily solved. I can't imagine you would be expected to solve something like this. Where does this problem come from, might I ask?

Also aren't Kepler's laws for planetary motion?

 

[gneill]

Also aren't Kepler's laws for planetary motion?

Yes, but you have a somewhat analogous problem here with an inverse-square central force. Really it's the Kepler Equation / Kepler Problem that would be pertinent rather than the three planetary laws themselves. It involves a numerical method to solve for the position versus time for a body in orbit (and a radial trajectory is a type of orbit!).

Can you let us know what level / class you're in? Have you studied any Calculus yet?

 

[afk94]

Yes, but you have a somewhat analogous problem here with an inverse-square central force. Really it's the Kepler Equation / Kepler Problem that would be pertinent rather than the three planetary laws themselves. It involves a numerical method to solve for the position versus time for a body in orbit (and a radial trajectory is a type of orbit!).

Can you let us know what level / class you're in? Have you studied any Calculus yet?

I'm taking the second semester of General Physics at the college level. It's Physics P202. I haven't taken calculus yet.

 

[gneill]

I'm taking the second semester of General Physics at the college level. It's Physics P202. I haven't taken calculus yet.

Okay. So it looks like an exact solution via differential equation is not going to happen. Some type of approximation will be required.

One thing you might consider is doing some successive approximations to close in on a reasonable value. For example, assuming that the acceleration is constant and equal to the initial acceleration at the starting point, calculate a distance value for the given time. Next, divide the time into two halves and find the distance for the half time, then recalculate the acceleration for that distance and do the second half of the time at that new acceleration. How does the total distance compare to the initial calculation? If the difference is large, cut the problem into smaller steps and repeat until you converge on a good value.

If you are familiar with programming you could just write a loop to do the work.

 

[afk94]

Okay. So it looks like an exact solution via differential equation is not going to happen. Some type of approximation will be required.

One thing you might consider is doing some successive approximations to close in on a reasonable value. For example, assuming that the acceleration is constant and equal to the initial acceleration at the starting point, calculate a distance value for the given time. Next, divide the time into two halves and find the distance for the half time, then recalculate the acceleration for that distance and do the second half of the time at that new acceleration. How does the total distance compare to the initial calculation? If the difference is large, cut the problem into smaller steps and repeat until you converge on a good value.

If you are familiar with programming you could just write a loop to do the work.

So basically find the acceleration at multiple times and then average them?

 

[gneill]

So basically find the acceleration at multiple times and then average them?

Not quite. You employ each value of acceleration to take a small portion of the journey. All the discrete steps add up to the final distance. At the end of each step you calculate the next value of acceleration to employ.

 

[afk94]

Ok so I just learned about magnetism and magnetic fields and and moving charges. the formula my professor gave me in class was Bp=μ/4∏ (Qa*Va x r (hat) / r^2
Can I somehow use that for Electric field instead?

 

[gneill]

Ok so I just learned about magnetism and magnetic fields and and moving charges. the formula my professor gave me in class was Bp=μ/4∏ (Qa*Va x r (hat) / r^2
Can I somehow use that for Electric field instead?

The force due to a magnetic field on a moving charge behaves differently than an electric field. While an electric field results in a force parallel to the field and is independent of the motion of the charge, a magnetic field produces a force which is proportional to the velocity of the charge and perpendicular to both the direction of motion and the field direction. For a charge at rest a magnetic field produces no force on the charge.

When both electric and magnetic fields are present at the same time, the net force that the charge feels is called the Lorentz Force. It's something worth looking up (if only so that you can recognize the formula and what it's about for now) because it will come up pretty often in your studies of electromagnetic forces.

 

[afk94]

Man, Physics is so interesting.