Calculating the total electric field from two charges

[goohu]

Homework Statement

see attached picture

Relevant Equations

E = kQ/R

a) Should be pretty straight forward, from the equation E = kQ/R , we see that scaling is simply 1/R.

b) Here is gets a bit trickier. We know that q acts as a source (E-field points outwards) and -q acts as a sink (E-field points inwards). If the distance is far away do we consider the Q1 and Q2 to be close enough that they cancel out each other? It doesn't seem to be correct in regards to the given answers.

c) -

The solution to the problems are:
a) 1/R
b) 1/(R^3)
c) 1/(R^2)

 

[BvU]

Hello @goohu , !

Can you check your relevant equation ?

For part b) you can check here

And what is the meaning of the minus sign for part c) ?

 

[goohu]

Hello, nice to be here.

Relevant equation should be E = kQ/R^2. I think c) should be solved using the same idea as in b) so i just put a "-" for no meaning really.

a) I just double checked the solutions; the answer in a) is actually 1/R^2. So I guess my reasoning was correct?

b) I checked the link but I don't really get the hint. What are reasonable assumptions to make? If we are looking at the E-field at a distance far away, how does the distance between the two charges matter?

What would be the difference if Q2 were placed at (0,1) or (1000, 1000) ?

c) I attached a picture of how the equipotential lines will look for two charges. In our case it will be reversed, since Q1 = q and Q2= -2q.

I guess what I don't understand is how the equipotential lines relates to the electric field strength.

 

[BvU]

Your picture isn't really relevant when the exercise asks for fields at great distance. What counts is the net charge, and the first order in r^-1 so for a) and c) you get the same answer.

b) is trickier: there net charge is zero and in the most sensitive direction (along the line connecting the charges -- think why that is so) the leading term is in r^-3

I checked the link but I don't really get the hint

move upward for V and differentiate to get E

 

[goohu]

Okay, I think I get it. Thanks for the explanation.

Is there a specific reason why you solve for the electric potential first in b) and then derivate it to get the electric field? Is this a special case of some form?

In a) and c) we don't need to do this since we only care about the net charge as stated, so in those cases we can just look at the equation for E-fields?

 

[BvU]

Is there a specific reason why they solve for the electric potential first in b) and then derivate it to get the electric field? Is this a special case of some form?

I suppose it's easier. You can find out by determining the ##\vec E## field directly

Special case ? ##\vec E =-\vec \nabla V## always (in electrostatics) !

 

[BvU]

how the equipotential lines relates to the electric field strength

Equipotential lines are like height lines on a detailed map: when they are close together, you know it's steep there.

 

[Joshy]

I don't know if I agree with any of the solutions. I don't think I do.

The question asks how does the electric field scale. The procedure I would do is solve for the electric field with only that first point charge ##Q_1##. I would normalize all solutions to that answer (divide the solution by this answer is what I'm really saying in plain English), and so if the electric field does not change at all, then it scales by ##1##. Just keep this in mind for after you solve the total electric field when there are two point charges.

After that, then I would place the second point charge and solve. What they're really testing you on is understanding superposition. It's a fancy way of saying that the total electrical field is the sum of the fields caused by individual components. You've already solved for the fields from ##Q_1##; so: you only need to solve for the fields from ##Q_2## and then add these two together. They've simplified the problem for you by making your observation point very far away, which suggests that ##R_1 \approx R_2##... for all practical purposes from far away the look about the same.

$$E_{\text{total}} = E_{Q1} + E_{Q2} = \left ( {1 \over {4\pi\epsilon}} {{q_1} \over {R_1^2}} + {1 \over {4\pi\epsilon}} {{q_2} \over {R_2^2}} \right ) \hat R$$

Just copying the equations straight from the textbooks. I would imagine you could do some simplification if ##R_1 \approx R_2## and then have an equation that has only one unknown parameter ##q_2##.

You're already in the right direction when ##Q_2 = 0## the fields do not change, with the total electric field being ##k { {q} \over {R^2} }##, and since that was the solution for the point charge by itself it has scaled by ##1##.

Alright. I think what I said here would make the professor unhappy.

 

[BvU]

The question asks how does the electric field scale.

It clearly says "How does the electric field scale with distance far away if .."

Which means an answer like "as r^-2" or "as r^-3" is sought.

[ Post edited slightly by a Mentor ]

 

[berkeman]

Thread locked temporarily for Moderation...

 

[berkeman]

Thread re-opened.

 

[Joshy]

Fair enough. I'm okay with admitting I was wrong and I thought the question was trying to demonstrate superposition and using the far field to simplify the problem. I was overly simplifying and going through it myself, and now I am convinced my earlier approach would not make the professor happy.

I was trying to go through the problem by summing up the E fields, which I don't think would be fundamentally wrong, but I came across some challenges that didn't make the far field approximation so useful. My textbook shows a suspiciously similar problem and starts by determining the voltage, which threw me off at first... because of the far field some of the variables could be very easily replaceable with the approximations. One of them of course was ##R_1 \approx R_2## they replaced ##R_1 R_2 \approx R^2##.

The other approximation was a bit tricky required using the law of cosine you'll get the vector ##\vec R_2 - \vec R_1 \approx dcos(\theta)## where ##d## is the distance between the two charges; you'll also have to assume that the lines are parallel to each other.

To convert back to the electric field you'll need to apply the equations BvU provided.

So far as the scaling thing goes... I'm still not sure. If you have time, then I would ask the professor; otherwise: You'll have to use your best judgement and justify your interpretation later if it was wrong.

You asked what happens if you move the second point to a different spot much further away from the first point charge. Could you still use the far field approximation and apply the same approach (or the same equations)? I think it would be a fun learning experiment try various observation points and reapply the equations or problems... compare it to the more direct approach summing the field components (you can't just add the fields that in direction ##\hat R_1## and ##\hat R_2## so you'll have to break it up into its ##\hat x##, ##\hat y##, and ##\hat z## components)

 

[goohu]

I suppose it's easier. You can find out by determining the ##\vec E## field directly

Special case ? ##\vec E =-\vec \nabla V## always (in electrostatics) !

Okay, then let me try.

Our total electric should be : Etot = kq(1/r1^2 - 1/r2^2), where r1 is the distance to the positive charge and r2 is the distance to the negative charge. Rewriting the equation: Etot = kq * (r2^2 - r1^2 )/r^4 , if we approximate r1^2*r2^2 to r^4.

Can we simplify it further to get the correct answer?

 

[BvU]

Our total electric should be : Etot = kq(1/r1^2 - 1/r2^2),

No. ##\vec E## is a vector with a direction and a magnitude, not just a magnitude.

 

[Joshy]

I was wrong earlier and wanted to try solving this more brute force way just to see (summing up the electric fields).

I think the professor was anticipating on people attempting to copy the textbook solution. What I've seen in the textbook has two charges sitting along the ##z## axis separated by a distance centered at the origin (the charges are equal to each other but one is the negative counterpart of the other). What's the solution in the textbook for that problem? In the ##\hat R## direction you'll have ##{ {2qdcos\theta} \over {4 \pi \epsilon R^3}}##. This ##\theta## is the angle between the observation point and the positive ##z## direction.

If you were to incorrectly continue with the method above (summing the fields without considering the direction), then you'd stumble upon what would look like a happy accident because it matches the textbook solution for the ##\hat R## direction, but there's a problem: The dipole in your problem is rotated relative to the drawing above (the other problem of course is you'd be completely ignoring the ##\hat \theta## direction). You could simplify ##R_2 - R_1## using the same approximation and say that ##R_2 + R_1 \approx 2R## to massage it into the textbook solution at least for the ##\hat R## direction.

Spoiler

Just a reminder that the solution or approach would be incorrect.

How to manage this problem if I don't know anything about the voltage or didn't want to do a derivative. You could still solve for this summing up the electric fields, but you'd have to do it much like other problems and you may have to do it for part (c) anyways.

How do we normally solve this for other textbook problems? You break it up into vector components and sum up the components. I'm aiming for the lazy approach here (maybe it's wrong so feel free to correct me I don't mind) and there's something very convenient about this problem. What I can do is call ##\theta## the angle between ##\vec R_1 - \vec R_2## and ##\vec R_1##; I could also say that ##\hat R_1## is the same direction as ##\hat R##. You've solved for this problem in part (a) so you're already done with half of it if you think about it this way.

How to break ##\vec R_2## into ##\hat R## components and ##\hat \theta##? Good news is ##\hat R_1## has no ##\hat \theta## components with this approach. I would take the projection of ##\vec R_1 - \vec R_2## onto ##\vec R_1##, then you know that the ##\hat R## component is ##R - dcos\theta##. The ##\hat \theta## component isn't too bad of a geometry problem ;).

When you do the mathematics you'll come across a seemingly shocking result if you assume that ##R \gg d## and you happen to know the answer from a textbook solution of a dipole where the two charges on the ##z## axis. If you rotate the textbook solution a bit... it'll give you confidence in the approach, and the results look reasonable too when you reuse the approach for ##Q_2 = -2q##.