Understanding Charge Density and Electric Fields

In summary: Im not sure how to do the math to find the electric field from the charge density. Thanks!The rod has a charge of -6.5uC/m, so the point is attracted to the rod with a field of -6.5uC/m.
  • #1
NBAJam100
146
0

Homework Statement



A point exists a distance D away from a rod of length L ( (point) |-----d-----| (ROD) ) So the point is in allignment with the rod, charged with a uniform charge density of -6.5uC/m

Homework Equations



E= Integral(k(dq/r^2)

The Attempt at a Solution



Im not asking for help on finding the answer, i just want to understand a bit more about charge density.

It says that it is (-), in short, does that mean that the rod has a negative charge? meaning that the point (positive ) will be attracted to the rod? Thanks.

One more thing, is the radius (r) the distance from the point to the center of the rod?
 
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  • #2
Hi NBAJam100,

NBAJam100 said:

Homework Statement



A point exists a distance D away from a rod of length L ( (point) |-----d-----| (ROD) ) So the point is in allignment with the rod, charged with a uniform charge density of -6.5uC/m

Homework Equations



E= Integral(k(dq/r^2)

The Attempt at a Solution



Im not asking for help on finding the answer, i just want to understand a bit more about charge density.

It says that it is (-), in short, does that mean that the rod has a negative charge? meaning that the point (positive ) will be attracted to the rod? Thanks.

If the charge density is negative, then the charge will be negative. (When the charge density is uniform, the linear charge density is just how much charge there is per unit length.)

One more thing, is the radius (r) the distance from the point to the center of the rod?

Do you mean the r in your integral? That r is the distance from the dq to the point where you're calculating the field at.

One way to understand the integral is to say we would like to break up the rod into an series of point charges, each with charge dq. The field from each of these charges dq is found from Coulomb's law (that's where the integrand comes from, and that's where the r comes in), and the integration adds up the fields from all of those charges.
 
  • #3
alphysicist said:
Hi NBAJam100,




Do you mean the r in your integral? That r is the distance from the dq to the point where you're calculating the field at.

One way to understand the integral is to say we would like to break up the rod into an series of point charges, each with charge dq. The field from each of these charges dq is found from Coulomb's law (that's where the integrand comes from, and that's where the r comes in), and the integration adds up the fields from all of those charges.

So if the r comes from the distance from the point to the dq, then does that mean that the r is not constant and has to be considered in the integral because there are many dq's on the rod (all of them make up Q). Or do you mean that I am measuring only one distance, from the point to that one sinlge dq that i am basing the problem off of? Thanks alph
 
  • #4
NBAJam100 said:
So if the r comes from the distance from the point to the dq, then does that mean that the r is not constant and has to be considered in the integral because there are many dq's on the rod (all of them make up Q). Or do you mean that I am measuring only one distance, from the point to that one sinlge dq that i am basing the problem off of? Thanks alph

The value r is not a constant and is part of the integration. You'll need to rewrite both dq and r in terms of an integration variable.

So, for example, if you decide to integrate from the left end to the right end of the rod, and let the integration variable x represent the distance from the left end, what would dq and r be?
 
  • #5
alphysicist said:
The value r is not a constant and is part of the integration. You'll need to rewrite both dq and r in terms of an integration variable.

So, for example, if you decide to integrate from the left end to the right end of the rod, and let the integration variable x represent the distance from the left end, what would dq and r be?


Hmmm... ok i kind of think i got what you are saying. Either that or this is totally wrong:

So i did E= k (integral) dq/x^2. Using the charge density (-6.5x10^-6 C/m) I found dq= (-6.5x10^-6 C/m)(dl). So i plugged that into the integral and pulled the charge density out as a constant. After integration i got E= k(-6.5x10^-6 C/m) (-1/x)... now i think this is right, but I am not sure what to plug into get the electric field. The rod is 1.5m long and the point is .5m away from the rod. So would i plug in 2m (the total distance from the end of the rod to the point) to get the electric field at the point? If that is the case, i found the answer to be 29,250 N/C
 
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  • #6
NBAJam100 said:
Hmmm... ok i kind of think i got what you are saying. Either that or this is totally wrong:

So i did E= k (integral) dq/x^2. Using the charge density (-6.5x10^-6 C/m) I found dq= (-6.5x10^-6 C/m)(dl). So i plugged that into the integral and pulled the charge density out as a constant. After integration i got E= k(-6.5x10^-6 C/m) (-1/x)... now i think this is right, but I am not sure what to plug into get the electric field. The rod is 1.5m long and the point is .5m away from the rod. So would i plug in 2m (the total distance from the end of the rod to the point) to get the electric field at the point? If that is the case, i found the answer to be 29,250 N/C

This will be a definite integral, so you need to determine the limits.

The integrand was Coulomb's law for each dq; so then what will be the limits of the integral? (Remember it will be different depending on how you defined your integation variable.)
 
  • #7
alphysicist said:
This will be a definite integral, so you need to determine the limits.

The integrand was Coulomb's law for each dq; so then what will be the limits of the integral? (Remember it will be different depending on how you defined your integation variable.)


well, from the information i have i don't think i can determine the area of the rod... but for some reason i feel that the area should be a factor... regardless, if my integration variable was x, (or l for distance of length). So wouldn't the bounds be 0-2m. Or, 0m= the end of the rod, 2m = the end of the rod to the point. So by my logic, my answer would be correct. (assuming my logic is correct).
 
  • #8
NBAJam100 said:
well, from the information i have i don't think i can determine the area of the rod... but for some reason i feel that the area should be a factor... regardless, if my integration variable was x, (or l for distance of length). So wouldn't the bounds be 0-2m. Or, 0m= the end of the rod, 2m = the end of the rod to the point. So by my logic, my answer would be correct. (assuming my logic is correct).

I don't know what the 2 meters corresponds to. I thought all we had was d=distance from point to rod, and L=length of rod?

But about the limits: In the last post you had the integrand as:

[tex]
k\lambda\frac{dx}{x^2}
[/tex]
where [itex]\lambda[/itex] is the linear charge density of the rod. (There are other ways to set up the integrand, but this is fine.) Now you want to integrate this, and to do that you need to decide on limits.

So your integration variable is x, and you want to integrate from the left end to the right end of the rod. At the left end, what is the value of x (in terms of d and L)? That will be the lower limit; in other words it is the value of the integration variable x that makes the integrand correct for the dq at the left end.

Then the upper limit will be the value of x will correspond to the right end of the rod.


About your post: I don't know what the 2 meters is. However, with the way you set up your integrand, your lower limit cannot be 0. Your denominator is x^2, so x is the distance from dq to the point. But the left end is not zero meters away, so the lower limit can't be 0. Do you see what your two limits have to be?
 
  • #9
alphysicist said:
I don't know what the 2 meters corresponds to. I thought all we had was d=distance from point to rod, and L=length of rod?

But about the limits: In the last post you had the integrand as:

[tex]
k\lambda\frac{dx}{x^2}
[/tex]
where [itex]\lambda[/itex] is the linear charge density of the rod. (There are other ways to set up the integrand, but this is fine.) Now you want to integrate this, and to do that you need to decide on limits.

So your integration variable is x, and you want to integrate from the left end to the right end of the rod. At the left end, what is the value of x (in terms of d and L)? That will be the lower limit; in other words it is the value of the integration variable x that makes the integrand correct for the dq at the left end.

Then the upper limit will be the value of x will correspond to the right end of the rod.


About your post: I don't know what the 2 meters is. However, with the way you set up your integrand, your lower limit cannot be 0. Your denominator is x^2, so x is the distance from dq to the point. But the left end is not zero meters away, so the lower limit can't be 0. Do you see what your two limits have to be?


Ok, so d= .5m (the point to the beginning of the rod), L of the rod = 1.5 meters. So, my lower bound would be .5 (or the distance from the point to the rod (d)), my upper bound would be 2m, the distance from the point to the end of the rod (d+L). So i would just plug those values into get the correct value...
 
  • #10
NBAJam100 said:
Ok, so d= .5m (the point to the beginning of the rod), L of the rod = 1.5 meters. So, my lower bound would be .5 (or the distance from the point to the rod (d)), my upper bound would be 2m, the distance from the point to the end of the rod (d+L). So i would just plug those values into get the correct value...

Those sound like the right integration limits to me. Did you get the right answer?
 
  • #11
alphysicist said:
Those sound like the right integration limits to me. Did you get the right answer?

For my final answer i got E= -87,750 N/C...

Thanks for all of the help alph! i really appreciate it!
 
  • #12
NBAJam100 said:
For my final answer i got E= -87,750 N/C...

Thanks for all of the help alph! i really appreciate it!

That seems like the right magnitude to me.

But you might need to be careful: what does the minus sign mean in your answer?

If you include a sign then it indicates the direction of this electric field; so for this case, which way is this electric field pointing at that point? And does that match with your definition of positive and negative directions?
 
  • #13
alphysicist said:
That seems like the right magnitude to me.

But you might need to be careful: what does the minus sign mean in your answer?

If you include a sign then it indicates the direction of this electric field; so for this case, which way is this electric field pointing at that point? And does that match with your definition of positive and negative directions?

I determined that the point was positive, and the rod was negatively charged (found by the charge density being negative). So if the point is positive, and the rod is negative, then the point is attracted to the rod. So the negative sign is right, the direction of the field from the point would be pointing towards the rod (or towards the negative (-)). Correct?
 
  • #14
NBAJam100 said:
I determined that the point was positive, and the rod was negatively charged (found by the charge density being negative). So if the point is positive, and the rod is negative, then the point is attracted to the rod. So the negative sign is right, the direction of the field from the point would be pointing towards the rod (or towards the negative (-)). Correct?

No, I don't believe that's correct. The positive or negative sign of the electric field will tell the direction--and that depends on the coordinate system. So if you set up your x-axis in the normal way with positive meaning to the right and negative meaning leftwards, then a negative value of the electric field would mean that the electric field is pointing to the left. But you said it's pointing toward the rod, which is to the right.



Think of it this way: you found the electric field at a point to the left of the rod, and you know the electric field there is pointing to the right (toward the rod).

Now suppose you next found the electric field at the opposite side of the rod (on the right), the same distance away. It would have the same strength of the electric field, and it's still pointing toward the rod, but now "toward the rod" means to the left. Since it points in an opposite direction, it would need to have the opposite sign of your original point. So the sign of the electric field is not the sign of the charge, it's simply indicates the direction of the field.



(Remember that what you are really doing with the sign is reporting the x-component of the electric field, because you already know it has to be in the x-direction.)
 
  • #15
alphysicist said:
No, I don't believe that's correct. The positive or negative sign of the electric field will tell the direction--and that depends on the coordinate system. So if you set up your x-axis in the normal way with positive meaning to the right and negative meaning leftwards, then a negative value of the electric field would mean that the electric field is pointing to the left. But you said it's pointing toward the rod, which is to the right.



Think of it this way: you found the electric field at a point to the left of the rod, and you know the electric field there is pointing to the right (toward the rod).

Now suppose you next found the electric field at the opposite side of the rod (on the right), the same distance away. It would have the same strength of the electric field, and it's still pointing toward the rod, but now "toward the rod" means to the left. Since it points in an opposite direction, it would need to have the opposite sign of your original point. So the sign of the electric field is not the sign of the charge, it's simply indicates the direction of the field.



(Remember that what you are really doing with the sign is reporting the x-component of the electric field, because you already know it has to be in the x-direction.)


Ok, so i THINK i get what this means... are you saying that i was finding the direction of the field from the left of the point... It was pretty much a given that from the right side of the point the field is facing towards the rod... but what that sign indicated was that the sign is in the negative direction from the left side of the point? If so, i understand that fully... So pretty much it would look like this:

---> = field lines

<---<---<---( + )--->--->---> ( ( - ) Charged rod) <---<---<---<

And then there would be other field lines curving from the outside of the positive going to the negative.
Once again, i can't thank you enough for being so patient with me alph.
 
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  • #16
NBAJam100 said:
Ok, so i THINK i get what this means... are you saying that i was finding the direction of the field from the left of the point... It was pretty much a given that from the right side of the point the field is facing towards the rod... but what that sign indicated was that the sign is in the negative direction from the left side of the point? If so, i understand that fully... So pretty much it would look like this:

---> = field lines

<---<---<---( + )--->--->---> ( ( - ) Charged rod) <---<---<---<

And then there would be other field lines curving from the outside of the positive going to the negative.
Once again, i can't thank you enough for being so patient with me alph.

I'm a bit confused about your diagram. There's not a charge at the point you are calculating is there? Usually they say they want you to find the electric field at a certain point; but there's nothing actually at that point. It's just empty space.

But even if there is a charge there (which it really does not sound like there is), if you're asking about the electric field of the rod, the answer will be the same whether there's a positive charge, negative charge, or nothing at that point. You are just asked what is the electric field from the rod at that point, right? So the diagram is:

--->--->---> ( ( - ) Charged rod) <---<---<---<-



By the way, I noticed you never explicitly stated the question. Usually these questions have two forms: 1. they ask for the magnitude of the field, in which case just the absolute value is needed; or 2. they want the magnitude and direction of the field, in which case the answer here would be "87750 N/C to the right" or "87750 N/C in the positive x-direction" or something similar.

Usually, if they wanted the sign of the number to indicate the direction, they would say that explicitly (something like "in your answer, let positive be to the right"). So if it's still not making sense, would you post the exact statement of the question?
 
  • #17
alphysicist said:
I'm a bit confused about your diagram. There's not a charge at the point you are calculating is there? Usually they say they want you to find the electric field at a certain point; but there's nothing actually at that point. It's just empty space.

But even if there is a charge there (which it really does not sound like there is), if you're asking about the electric field of the rod, the answer will be the same whether there's a positive charge, negative charge, or nothing at that point. You are just asked what is the electric field from the rod at that point, right? So the diagram is:

--->--->---> ( ( - ) Charged rod) <---<---<---<-



By the way, I noticed you never explicitly stated the question. Usually these questions have two forms: 1. they ask for the magnitude of the field, in which case just the absolute value is needed; or 2. they want the magnitude and direction of the field, in which case the answer here would be "87750 N/C to the right" or "87750 N/C in the positive x-direction" or something similar.

Usually, if they wanted the sign of the number to indicate the direction, they would say that explicitly (something like "in your answer, let positive be to the right"). So if it's still not making sense, would you post the exact statement of the question?

Ok, I am getting two things mixed up i think haha. You're right, there is no charge there, the question is this: "Find the magnitude and direction of the net electric field at point P, a distance of .5m away from the end of a rod of length 1.5m charged with a uniform charge density of -6.5 uC/m." I understand how to get the answer, i guess i was just confused about the negative in my answer with regards to the direction of the field.
 
  • #18
NBAJam100 said:
Ok, I am getting two things mixed up i think haha. You're right, there is no charge there, the question is this: "Find the magnitude and direction of the net electric field at point P, a distance of .5m away from the end of a rod of length 1.5m charged with a uniform charge density of -6.5 uC/m." I understand how to get the answer, i guess i was just confused about the negative in my answer with regards to the direction of the field.

Oh, okay. So the magnitude would be just the 87750 N/C, and the direction would be to the right, like we had a few posts back. Looks great!
 
  • #19
Thank a million alph! i must say i learned a hugeee amount from these posts!
 
  • #20
Sure, glad to help!
 

FAQ: Understanding Charge Density and Electric Fields

What is an electric field?

An electric field is a physical quantity that describes the influence that a charged particle has on other charged particles in its vicinity. It is a vector quantity, meaning it has both magnitude and direction. Electric fields are created by charged particles and can exert a force on other charged particles.

How is an electric field calculated?

An electric field is calculated by dividing the force exerted on a charge by the magnitude of the charge. This can be expressed mathematically as E = F/q, where E is the electric field, F is the force exerted on the charge, and q is the magnitude of the charge.

What is the SI unit for electric field?

The SI unit for electric field is newtons per coulomb (N/C). This unit represents the force exerted on a charge of one coulomb in an electric field.

What is the difference between an electric field and an electric potential?

An electric field describes the force experienced by a charged particle in an electric field, while electric potential describes the energy possessed by a charged particle in an electric field. In other words, electric potential is the potential energy per unit charge at a given point in an electric field.

How are electric fields used in everyday life?

Electric fields are used in many everyday devices, such as batteries, electric motors, and generators. They are also used in the transmission and distribution of electricity through power lines. Additionally, electric fields are used in medical equipment such as MRI machines and in electronic devices like computers and smartphones.

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