Electric field due to two charges PLEASE HELP!

[stunner5000pt]

Find the point (or points) where the electric field due to two charges -5q and +2q is zero and separated by a distance a metres.

-5q ------------------- +2q

WEll first of i know that there is a point that is along the axis on the right side closer to 2q but farther from the negative.

But i'm wondering if there is a point somewhere else is space (other than infinity) where the electric field is zero.

Since the electric field due to the negative is pointing towards it and positive points outward i wonder if there is a point where the two vectors cancel out anywhere else is space.

Am i right in assuming that there is no other point besides the point on the axis??

[stunner5000pt]

Which axis now? Since the field moves from positive to negative charge there's no place in between the two and moves toward negative and away from positive not in between the two so ... no there's no other point although symmetry makes the y component 0 along the axis.

i mean the axis upon which they are mounted

so there is a point along this axis where E = 0 , however, no other point exists?

[stunner5000pt]

There are no points along the axis any closer than infinitely far away that would have E = 0. This is only achieved by two positive or two negative charges.

think about how the field is flowing:

----------------> -q <--------------------- +q ------------------------>

but how about a point t such that 2kq/t^2 - 5kq/(a+t)^2 = 0 and i can solve for t and it is a positive number

[cepheid]

Hmm...I'm taking electromagnetism right now, so I wanted to make sure I understood why he is correct. Call the left charge q1, the right q2:

\vec{E}_1 = \frac{1}{4\pi\epsilon_0}\frac{q_1}{r_1^2}\hat{r}_1

\vec{E}_2 = \frac{1}{4\pi\epsilon_0}\frac{q_2}{r_2^2}\hat{r}_2

r_1 and r_2 are the magnitudes of the separation between the field point and the first and second source points (respectively)

\hat{r_1}\ \ \text{and}\ \ \hat{r_2}

are unit vectors pointing in the direction from each source point to the field point. By the principle of superposition, the net electric field is the vector sum of the two, and this sum can only be zero for some point at which:

\hat{r_1} = -\hat{r_2}

and this is only true along the line of action between the two charges. For any other point, the vector sum will yield some resultant electric field vector.

EDIT: Never mind, I'm wrong. I was thinking about two like charges. I just looked at vsages diagram, and it seems that \hat{r_1} = \hat{r_2} along the line of action between them i.e. the fields due to each one always point in the same direction. Thanks for clearing this up.

[stunner5000pt]

So there's no point?

You're right both the fields almost reinforce each other with direction, one points away one towards it, so this maenas there exists no point???!

[stunner5000pt]

No point with finite coordinates, no. If you would like there is a set of points that doesn't really have a particular shape at the infinities that would have a field equal to zero but the shape of this set of points is unclear. It may be circular, square, etc I wouldn't worry about it.

sweet

thank you very much

i am in your debt

[nrqed]

So there's no point?

You're right both the fields almost reinforce each other with direction, one points away one towards it, so this maenas there exists no point???!

Stunner500,

Of course, there is a point (other than at infinity) where the two E fields cancel out to give a total E field equal to zero. It is to the right of the 2q (as I think you pointed out in your first post). I am not sure what vsage meant (maybe he/she was just focusing on points between th echarges?? I don't know).

But other than that point (which you can find by setting the two magnitudes equal), there is no other point with E_total=0 (other than at infinity)

Pat

[stunner5000pt]

Stunner500,

Of course, there is a point (other than at infinity) where the two E fields cancel out to give a total E field equal to zero. It is to the right of the 2q (as I think you pointed out in your first post). I am not sure what vsage meant (maybe he/she was just focusing on points between th echarges?? I don't know).

But other than that point (which you can find by setting the two magnitudes equal), there is no other point with E_total=0 (other than at infinity)

Pat

ok im wondering who's the superior authority on this

that's what i thought in the first place

[nrqed]

No point with finite coordinates, no. If you would like there is a set of points that doesn't really have a particular shape at the infinities that would have a field equal to zero but the shape of this set of points is unclear. It may be circular, square, etc I wouldn't worry about it.


The point located at a distance {{\sqrt 2} \over {\sqrt 5} - {\sqrt 2}} a \approx 1.72 a to the right of the 2q has a total E field equal to zero.


Pat

[stunner5000pt]

The point located at a distance {{\sqrt 2} \over {\sqrt 5} - {\sqrt 2}} a \approx 1.72 a to the right of the 2q has a total E field equal to zero.


Pat

but what about the direction

you would put the two equal to each other

2kq / t^2 = -5kq / (a+t)^2

or would they bne the same signs?

[nrqed]

ok im wondering who's the superior authority on this

that's what i thought in the first place


I am glad you read my post because I was worried you would be left with the wrong idea!! (btw, I give the exact position in another posting).

As fo the superior authority, I can give you my credentials (PhD in phys, 8 years of college level teaching) but in the end, it must be decided by physics and maths. If you pick a point to the right of the 2q, the two E fields point in opposite directions. And because you are closer to the weaker charge (in absolute value), there is necessarily a point where the two magnitudes will be equal. Call "x" the distance between that point and the 2q charge. Then you only need to solve 2q/x^2 = 5q/(x+a)^2 (I used the absolute values of the charges since I am imposing the two magnitudes to be equal). Solving for x gives the answer I presented in my other post.

Pat

[stunner5000pt]

PhD in phys, 8 years of college level teaching




i'm convinced, thank you

[vsage]

Hrm I am going to bring this up at my next physics lecture. I'm certainly open to ideas but I just hadn't thought of it as possible before. Sorry stunner! I would go with Nrq definitely. I am a first year college student (engineering major) so Nrq has quite a bit of education on me.

[stunner5000pt]

no problem

[nrqed]

but what about the direction

you would put the two equal to each other

2kq / t^2 = -5kq / (a+t)^2

or would they bne the same signs?

This type of problem is better done in two steps. In the first step, you worry about the directions only: you look for points where the two E field produced by the two charges are in opposite directions. In this case, it's easy to see that any point to the left of the -5q or to the right of the +2q satisfy this (if the charges had been of equal signs, you would have picked a point somewhere between the charges).

Now, you consider the magnitudes. The vectors must not only be pointing in opposite directions, they must have the same magnitude. This rules out the points on the left of the 5q (because these points are always closer to the stronger charge (in absolute value) so there is no way for the two magnitudes to be equal). That leaves the points to the right of the 2q.

Now, you impose that the two *magnitudes* be equal, which means you impose

{ |-5 q| \over (x+a)^2 } = { |2 q| \over x^2}

where my vertical bars denote the absolute values. And then you solve.

Basically, what we did is to solve the condition \vec E_1 + \vec E_2 = 0 or \vec E_1 = - \vec E_2 which means teh two vectors must be opposite and of equal magnitude. We took care of the directions first and the when came the time of finding the exact position, we work with the magnitudes.

I hope this clarifies things for you.

Pat

[stunner5000pt]

This type of problem is better done in two steps. In the first step, you worry about the directions only: you look for points where the two E field produced by the two charges are in opposite directions. In this case, it's easy to see that any point to the left of the -5q or to the right of the +2q satisfy this (if the charges had been of equal signs, you would have picked a point somewhere between the charges).

Now, you consider the magnitudes. The vectors must not only be pointing in opposite directions, they must have the same magnitude. This rules out the points on the left of the 5q (because these points are always closer to the stronger charge (in absolute value) so there is no way for the two magnitudes to be equal). That leaves the points to the right of the 2q.

Now, you impose that the two *magnitudes* be equal, which means you impose

{ |-5 q| \over (x+a)^2 } = { |2 q| \over x^2}

where my vertical bars denote the absolute values. And then you solve.

Basically, what we did is to solve the condition \vec E_1 + \vec E_2 = 0 or \vec E_1 = - \vec E_2 which means teh two vectors must be opposite and of equal magnitude. We took care of the directions first and the when came the time of finding the exact position, we work with the magnitudes.

I hope this clarifies things for you.

Pat

While we're on this topic can you assist me in something similar to this

ill post in a new thread though