# Electric Fields - continuous charge distributions

1. Aug 31, 2014

### Xgens

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

3. 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: Aug 31, 2014
2. Aug 31, 2014

### Simon Bridge

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.

3. Aug 31, 2014

### Xgens

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.

4. Aug 31, 2014

### Simon Bridge

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.

5. Sep 1, 2014

### Xgens

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!

6. Sep 1, 2014

### Simon Bridge

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.

7. Sep 1, 2014

### Xgens

I see, alright let me work it out and i will post it. Thanks!

8. Sep 1, 2014

### Xgens

This is what i tried so far

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9. Sep 1, 2014

### Simon Bridge

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. Sep 1, 2014

### Xgens

It's on the question 1b and 1c that are related to this.

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• ###### soln_P2212sp2012_act02_electricfields-3.pdf
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11. Sep 1, 2014

### Simon Bridge

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?