Time to learn and understand my mistakes.

[flyingpig]

Homework Statement

If I get lucky, I'll average out 80% on my final grade in E&M this term.

ANyways I have a question

Let's say I have a spherical insulator of charge +Q and radius R and at a distance 2R measured from its center exists a particle with charge -Q.

What is the speed when this particle hits the surface of this insulator?

The Attempt at a Solution

I have two solutions, I know one of them is wrong...

\int \vec{F_e} \cdot \vec{dx} = \Delta K

Now, the problem is figuring out the limits of integration and I think this is where I went wrong

I had initally

\int_{2R}^{R} \frac{k_eQ(-Q)}{r^2} dr = \Delta K

But I realize this is wrong right? Because the convention is to bring a charge from infinity to the surface.

So it should be

\int_{\infty}^{R} \frac{k_eQ(-Q)}{r^2} dr = \Delta K

Am I right?

 

[Doc Al]

No. You're moving the charge from r = 2R to r = R, not from infinity. (In defining the potential energy at some point, you'll bring the charge in from infinity. Here you are interested in the change in potential energy.)

 

[cupid.callin]

why don't you use energy conservation?

 

[cupid.callin]

I have two solutions, I know one of them is wrong...

\int \vec{F_e} \cdot \vec{dx} = \Delta K

Now, the problem is figuring out the limits of integration and I think this is where I went wrong

I had initally

\int_{2R}^{R} \frac{k_eQ(-Q)}{r^2} dr = \Delta K

But I realize this is wrong right? Because the convention is to bring a charge from infinity to the surface.

So it should be

\int_{\infty}^{R} \frac{k_eQ(-Q)}{r^2} dr = \Delta K

Am I right?

Its just Work Energy Theorem ... it can have any limits ...

 

[flyingpig]

No. You're moving the charge from r = 2R to r = R, not from infinity. (In defining the potential energy at some point, you'll bring the charge in from infinity. Here you are interested in the change in potential energy.)

Wait, so I was initially right? This was a question on my final exam lol

 

[flyingpig]

why don't you use energy conservation?

Because I find it difficult to use, I thought what I am doing right now is the conversation conservation of energy (I keep spelling conservation wrong lol)

 

[Doc Al]

Wait, so I was initially right? This was a question on my final exam lol

Your initial idea of integrating from 2R to R was correct.

 

[cupid.callin]

Energy conservation is quite simple ...

here just PE will decrease and it will produce KE ...

so just write eqn for that :)

 

[Doc Al]

Because I find it difficult to use, I thought what I am doing right now is the conversation conservation of energy (I keep spelling conservation wrong lol)

What you're doing is equivalent to using conservation of energy, only 'harder' since you have to do the integration yourself.

 

[flyingpig]

What you're doing is equivalent to using conservation of energy, only 'harder' since you have to do the integration yourself.

I think it's easier lol

Can you guys tell me why I am wrong though when I second guessed myself?

I set the lower limit to infinity as to choose my "nullpoint".

Like you start from 0 and then you go to R

R is the distance between 2R and R (2R - R = R, so it you started from 3R, it would have been r = infinity to r = 2R)

 

[Doc Al]

Can you guys tell me why I am wrong though when I second guessed myself?

To find the work done, you integrate the force over the distance traveled, which is from 2R to R, not from infinity.

If you want to go from infinity, you'll need to integrate twice: From infinity to 2R and from infinity to R. Then subtract to find the work done from 2R to R. Doesn't much make sense.

 

[flyingpig]

Doc Al or cupid, I know I am hijacking my own thread, but this brings me to two good questions

Let's say (I guess in this case!), the energy initially associated is

\int \frac{k_eQ(-Q)}{r^2} dr = \frac{kQQ}{r} Where I am only concern with the definite integral right now.

Also, one of the charges is negative and one of them is positive

Now the result gives me a +U, what exactly does this mean? I know a gain in potential energy means a gain in Kinetic energy, but what other meaning is there in just terms of potential energy. What does it mean to have a decrease/increase in potential energy? Like what is it doing

 

[Doc Al]

Now the result gives me a +U, what exactly does this mean?

You are calculating the work done by the field when the charge moves from 2R to R, which is positive. The kinetic energy increases since the charges attract. The change in potential energy is the negative of that.

I know a gain in potential energy means a gain in Kinetic energy, but what other meaning is there in just terms of potential energy.

Assuming energy is conserved, a gain in potential energy means a decrease in kinetic energy.

What does it mean to have a decrease/increase in potential energy?

It's just a way of keeping track of the work you have to do against the field when a charge is moved.

 

[flyingpig]

Sorry I mean to say a "loss in KE", my hands and my brain, they don't like each other...

 

[flyingpig]

It's just a way of keeping track of the work you have to do against the field when a charge is moved.

But is it a positive or negative charge?>

 

[Doc Al]

But is it a positive or negative charge?>

The potential--not the potential energy--is defined using a positive unit charge.

The potential at a distance R from a point charge Q is kQ/R. The potential energy when a charge q is placed at R is kQq/R. If Q and q are oppositely charged, then the potential energy increases with distance.

 

[flyingpig]

Sorry for hijacking my own thread, but

[PLAIN]http://img832.imageshack.us/img832/17/unledqp.png

Initally the energy is negative, why?

U=\frac{-kqq}{r}

But one charge is negative and one is positive, so

U=\frac{kQQ}{L + l + R_1 + R_2}

BUt they have negative potential energy

 

[Doc Al]

U=\frac{-kqq}{r}

That's the potential energy between two charges, one positive (+q) and one negative (-q). It already incorporates the sign of the charges; q is just the magnitude of the charges. (See my last post for the general formula.)

Rather than rely on formulas, think it through. Since the charges attract, the potential energy must increase as they are pulled apart. And since the potential energy goes to zero at infinity, it must be negative for finite distances.

 

[flyingpig]

Doc Al, I thought of the perfect way of asking my qusetion from before why I thoughti t was from infinity to R.

Going back to mechanics. Let's say you are standing on the ground at x = 0 (x is horizontal) and you throw something up and then the object comes down.

Now my question is what the work done by gravity as it reaches the peak of the thrown.

Now I would just do

\int_{\infty}^{h_{max}} \vec{F_g} \cdot \vec{dr}

I took infinity because it can "set" it to 0, but the change in distance is still just h_max

 

[Doc Al]

Now I would just do

\int_{\infty}^{h_{max}} \vec{F_g} \cdot \vec{dr}

I took infinity because it can "set" it to 0, but the change in distance is still just h_max

Why do you keep dragging infinity into this? To find the work done by a force moving from one position to another, just use:

\int_{r_1}^{r_2} \vec{F_g} \cdot \vec{dr}

 

[flyingpig]

Why do you keep dragging infinity into this? To find the work done by a force moving from one position to another, just use:

\int_{r_1}^{r_2} \vec{F_g} \cdot \vec{dr}

Let's say you are on the ground and yuo throw a ball up. What is the work done by gravity as it reaches the highest point?

\int \vec{F_g} \cdot \vec{dr}


\int \frac{GMm}{r^2} dr

-\frac{GMm}{r}

It would make sense to put infinity for r so that the initial potential enregy is 0 and then put hmax for r

 

[Doc Al]

It would make sense to put infinity for r so that the initial potential enregy is 0 and then put hmax for r

Why do you think that would give you the work done from h=0 to hmax? (It would give you the work done from r = infinity to r = hmax, which is not what you want.)

 

[flyingpig]

Because you can do the "1/infinity" thing... and then say "uhh it is 0"

 

[Doc Al]

Now I would just do

\int_{\infty}^{h_{max}} \vec{F_g} \cdot \vec{dr}

I took infinity because it can "set" it to 0, but the change in distance is still just h_max

This will give you the work done from infinity to h_max, which is not the same as the work done from h_1 (ground) to h_max.

 

[flyingpig]

But the ground is h = 0, you can't divide by 0

 

[Doc Al]

But the ground is h = 0, you can't divide by 0

Don't confuse h, the height above the ground, with r, the distance between the mass and the center of the earth.

 

[flyingpig]

Can you do a numerical example for me?

 

[Doc Al]

Can you do a numerical example for me?

Why don't you do one? Try this: Lift a 1 kg mass 1 meter. How much work does gravity do during that motion?

h_initial = 0 (at the Earth's surface)
h_final = 1

To figure out the work using Newton's law of gravity, you'd have:
r_initial = radius of earth
r_final = radius of Earth + 1 m

Try doing it your way, integrating from r = infinity and you'll see it won't make much sense.

(Note: You're confusing the definition of gravitational PE, which is based on work done going from r = infinity to some point, and the work done to move something from one point to another. Of course you can use gravitational PE to easily figure out that work without doing any integration--it's already done for you--but that doesn't mean they are identical.)

 

[flyingpig]

Why don't you do one? Try this: Lift a 1 kg mass 1 meter. How much work does gravity do during that motion?

h_initial = 0 (at the Earth's surface)
h_final = 1

To figure out the work using Newton's law of gravity, you'd have:
r_initial = radius of earth
r_final = radius of Earth + 1 m

Try doing it your way, integrating from r = infinity and you'll see it won't make much sense.

(Note: You're confusing the definition of gravitational PE, which is based on work done going from r = infinity to some point, and the work done to move something from one point to another. Of course you can use gravitational PE to easily figure out that work without doing any integration--it's already done for you--but that doesn't mean they are identical.)

But if you are integrating from the radius of the Earth to radius of Earth + 1m, isn't that saying you are on the surface of the planet and you are throwing it 1m in space?? ( i know taht sounds silly...)

 

[Doc Al]

But if you are integrating from the radius of the Earth to radius of Earth + 1m, isn't that saying you are on the surface of the planet and you are throwing it 1m in space?? ( i know taht sounds silly...)

Yes, that's about right. I thought that was pretty close to what you had in mind:

Let's say you are standing on the ground at x = 0 (x is horizontal) and you throw something up and then the object comes down.

Now my question is what the work done by gravity as it reaches the peak of the thrown.

 

[flyingpig]

But the object isn't in space...it's in the sky. Which is less than the Earth's Radius.

 

[Doc Al]

But the object isn't in space...it's in the sky. Which is less than the Earth's Radius.

Note that the distance 'r' in the Newton formula for gravity is the distance from the object to the center of the earth. For something above the Earth's surface, that distance will be greater than the Earth's radius.

 

[flyingpig]

But the Earth's surface is the sky and the object can't always be higher than the sky

 

[Doc Al]

But the Earth's surface is the sky and the object can't always be higher than the sky

The Earth's surface is below the sky.

 

[flyingpig]

Then what is the real radius of the earth?