Electric Fields - continuous charge distributions

AI Thread Summary
To determine the net electric field at point (0,6) m due to a negatively charged plastic rod, the same mathematical approach used for positive charges applies, but with careful attention to the sign of the charge. The electric field direction will be towards the rod, as a positive test charge would be attracted to the negative charge distribution. It's important to set up the electric field vectors correctly and track the negative signs throughout the calculations. The general formula for the electric field can be derived before substituting numerical values, ensuring clarity in the final results. Understanding these principles allows for accurate computation of the electric field from continuous charge distributions.
Xgens
Messages
8
Reaction score
0

Homework Statement


A plastic rod of finite length carries an uniform linear charge Q = -5 μC along the x-axis with the left edge of the rod at the origin (0,0) and its right edge at (8,0) m. All distances are measured in meters.

Determine the magnitude and direction of the net electric field at a point (0,6) m, along the positive y-axis.

2. Relevant Equation

λ= Q/L
∫dx/(x^2+a^2)^3/2 = x/a^2(x^2+a^2)^1/2
∫xdx/(x^2+a^2)^3/2 = -1/(x^2+a^2)^1/2

The Attempt at a Solution



If the uniform linear charge Q is positively charge then I can solve this problem but since this is negatively charge, how do I go about solving this? I am referring to the electric field, if Q is positively charged then the electric field is pointing to northwest of point P. Thanks in advance.
 
Last edited:
Physics news on Phys.org
You solve it exactly the same way as for a positive charge. If it helps you feel better, try substituting Q=-q: q>0, when Q<0.

The direction of the electric field vector at a position is the direction of the force experienced by a small positive test charge placed there. So what happens if the field is produced by a negative charge distribution, compared with that for a positive charge distribution? How about this: compare the field from a positive point charge with that from a negative one.
 
Simon Bridge said:
You solve it exactly the same way as for a positive charge. If it helps you feel better, try substituting Q=-q: q>0, when Q<0.

The direction of the electric field vector at a position is the direction of the force experienced by a small positive test charge placed there. So what happens if the field is produced by a negative charge distribution, compared with that for a positive charge distribution? How about this: compare the field from a positive point charge with that from a negative one.

The direction of the electric field vector at a position is the direction of the force experienced by a small positive test charge placed there

*I understand this part, that's why I said electric field will be northwest due to point P.

So what happens if the field is produced by a negative charge distribution, compared with that for a positive charge distribution?

Well the test charge is going to attract to the field produced by a negative charge distribution. My problem lies in this part, since it's attract to the negative charge distribution throughout x-axis from 0m to 8m. Does that mean I have to draw my dE pierced through the negative charge distribution?

Thanks a lot for your help so far.
 
No - you set up the vectors at the start of the calculation and keep track of the minus signs.
The actual calculation is exactly the same for positive and negative charge distributions.

It sounds like there is a specific example that is stumping you.
 
  • Like
Likes 1 person
Simon Bridge said:
No - you set up the vectors at the start of the calculation and keep track of the minus signs.
The actual calculation is exactly the same for positive and negative charge distributions.

It sounds like there is a specific example that is stumping you.

Basically I can solve this problem with positive charge distribution but just keep track of the minus sign? If so, then now i know how to solve it. Thanks a lot for your help!
 
No - you do the algebra (and general working out) with the regular symbols, and you plug the numerical values in last, like normal. It's just that, this time, the numerical value for Q is negative. So put that negative number in at the end.

I don't see the problem here - please provide an example to illustrate where you get stuck.
 
I see, alright let me work it out and i will post it. Thanks!
 
This is what i tried so far
 

Attachments

  • 20140901_103030.jpg
    20140901_103030.jpg
    25.2 KB · Views: 550
  • 20140901_103041.jpg
    20140901_103041.jpg
    14.8 KB · Views: 528
That looks like a wall of maths.
Without the example to go with it I cannot see what you've done.
When you teacher or text gives you an example, they set it up for you too don't they? You know - describe the situation before they show you the maths?
 
  • #10
It's on the question 1b and 1c that are related to this.
 

Attachments

  • #11
OK - in that example, a finite rod carries a charge of +50μC
Ah I see - the example working is done entirely in magnitudes - so if λ<0 then |λ|>0 and you get the same answer.

Try like this:

$$d\vec E = \frac{k\lambda\;dx}{x^2+y^2} \frac{y\hat\jmath - x\hat\imath}{\sqrt{x^2+y^2}}$$ ... this is because ##\vec r = y\hat\jmath - x\hat\imath## is the vector pointing from the charge to the location.

The model answer also substituted the numbers in too soon ... the best practise is to work out the general relation and then plug the numbers in. So derive ##\vec E(y)## and then put ##y=10## at the end.

Note: in that setup, if a positive test charge would be repelled up and to the left, then changing the sign of the charge distribution just results in a positive test charge being attracted down and to the right. I think this was your intuition?
 
Back
Top