Solved: Electric Field of 2 Charged Particles on x-Axis

[lha08]

Homework Statement

Two charged particles are located on the x axis. The first is a charge +Q at x= -a. The second is an unknown charge located at x= +3a. The net electric field these charges produce at the origin has a magnitude of 2kQ/a^2. Explain how many values are possible for the unknown charge and find the possible values.

Homework Equations

The Attempt at a Solution

So what I did was that I first assumed that q is positive then if q is negative. For when I assumed it was positive, I got 2kQ/a^2= kQ/a^2 - kq/9a^2 from the formula for the electric field (I just put two equations together) and got -9Q=q. Then I tried it for when q is negative and got 9Q=k...

The problem is that the answers says that 'the field at the origin can be to the right if the unknown charge is -9Q, or the field can be to the left if the unknown charge is +27Q...'
I'm not sure if the method I used is right...because I can't seem to get +27Q...

 

[collinsmark]

This gets kind of tricky since we're dealing with vectors.

Let me walk you through what you've already done. Use this as an example to get to the second part.

In general we have (go ahead and change this to your own notational conventions if you have to)

\vec E = k \frac{Q_1}{r_1^2} \hat{r_1} + k \frac{Q_2}{r_2^2} \hat{r_2}

where \hat r_1 and \hat r_2 are unit vectors, pointing from the respective charge. It's these little pesky unit vector things that make adding up vector fields tricky. (More later about why working with potential makes things easier.)

Fortunately for this problem, everything is along the x-axis. Since Q₁ is at x = -a, its unit vector \hat r_1 is pointing in the positive x direction with respect to the test point (at the origin). So in this case we can say \hat r_1 = \hat x.

Q₂ is at x = 3, so it unit vector \hat r_2 is pointing in the negative x direction with respect to the test point (at the origin). So in this case we can say \hat r_2 = - \hat x.

So now, noting that for our test point at the origin, |r₁| = a, |r₂| = 3a, Q₁ = Q, and electric field at the origin is |E| = 2kQ/a²,

2k\frac{Q}{a^2} \hat r = k \frac{Q}{a^2} \hat{x} - k \frac{Q_2}{9a^2} \hat{x}

But we don't know whether \hat r points in the positive x direction or the negative x direction. So we'll have to try it both ways. If \hat r points in the positive x direction, the field is positive. And it's negative if \hat r points in the negative x direction.

\pm 2k\frac{Q}{a^2} \hat x = k \frac{Q}{a^2} \hat{x} - k \frac{Q_2}{9a^2} \hat{x}

And since we're all on the same positive pointing x-axis now (all vectors have the same unit vector), we can remove the pesky unit vector \hat x and just work with the magnitudes (ah, finally!).

\pm 2k\frac{Q}{a^2} = k \frac{Q}{a^2} - k \frac{Q_2}{9a^2}

A little more quick algebra can easily be done to simply things,

\pm 2Q = Q - \frac{Q_2}{9}

and simplifying a little more (I'm skipping some steps, you should work through this yourself, btw),

Q_2 = 9Q(1 \mp 2)

You've already solved one of the answers, -9Q. The other answer shouldn't be too tough.

As you can see, some of the biggest trouble in this problem was dealing with those pesky unit vectors. if you continue to study electrostatics you'll be introduced to something called electric potential. Electric potential is related to electric fields though integration. Electric potentials are scalar fields, so they don't have any of those unit vectors to complicate things. Then after summing up things in scalar land, one can get back to the electric field by taking the gradient. You have to do calculus to use the potentials, but it's often worth it.

 

[lha08]

This gets kind of tricky since we're dealing with vectors.

Let me walk you through what you've already done. Use this as an example to get to the second part.

In general we have (go ahead and change this to your own notational conventions if you have to)

\vec E = k \frac{Q_1}{r_1^2} \hat{r_1} + k \frac{Q_2}{r_2^2} \hat{r_2}

where \hat r_1 and \hat r_2 are unit vectors, pointing from the respective charge. It's these little pesky unit vector things that make adding up vector fields tricky. (More later about why working with potential makes things easier.)

Fortunately for this problem, everything is along the x-axis. Since Q₁ is at x = -a, its unit vector \hat r_1 is pointing in the positive x direction with respect to the test point (at the origin). So in this case we can say \hat r_1 = \hat x.

Q₂ is at x = 3, so it unit vector \hat r_2 is pointing in the negative x direction with respect to the test point (at the origin). So in this case we can say \hat r_2 = - \hat x.

So now, noting that for our test point at the origin, |r₁| = a, |r₂| = 3a, Q₁ = Q, and electric field at the origin is |E| = 2kQ/a²,

2k\frac{Q}{a^2} \hat r = k \frac{Q}{a^2} \hat{x} - k \frac{Q_2}{9a^2} \hat{x}

But we don't know whether \hat r points in the positive x direction or the negative x direction. So we'll have to try it both ways. If \hat r points in the positive x direction, the field is positive. And it's negative if \hat r points in the negative x direction.

\pm 2k\frac{Q}{a^2} \hat x = k \frac{Q}{a^2} \hat{x} - k \frac{Q_2}{9a^2} \hat{x}

And since we're all on the same positive pointing x-axis now (all vectors have the same unit vector), we can remove the pesky unit vector \hat x and just work with the magnitudes (ah, finally!).

\pm 2k\frac{Q}{a^2} = k \frac{Q}{a^2} - k \frac{Q_2}{9a^2}

A little more quick algebra can easily be done to simply things,

\pm 2Q = Q - \frac{Q_2}{9}

and simplifying a little more (I'm skipping some steps, you should work through this yourself, btw),

Q_2 = 9Q(1 \mp 2)

You've already solved one of the answers, -9Q. The other answer shouldn't be too tough.

As you can see, some of the biggest trouble in this problem was dealing with those pesky unit vectors. if you continue to study electrostatics you'll be introduced to something called electric potential. Electric potential is related to electric fields though integration. Electric potentials are scalar fields, so they don't have any of those unit vectors to complicate things. Then after summing up things in scalar land, one can get back to the electric field by taking the gradient. You have to do calculus to use the potentials, but it's often worth it.

Thanks a lot for answering, you were really helpful.
I have one more question, when you say unit vectors, are you referring to the electric field vector of each charge? Is a unit charge a vector that simply points towards another charge?

 

[collinsmark]

Thanks a lot for answering, you were really helpful.
I have one more question, when you say unit vectors, are you referring to the electric field vector of each charge? Is a unit charge something completely different?

A unit vector is simply something that points in a specific direction. The magnitude of a unit vector is always 1. But in general, a unit vector's direction can point anywhere.

In the exercise we worked out here, the unit vector \hat x always points in the positive x direction, with a magnitude of 1. Similarly, \hat y points in the y direction. But when dealing with point charges, the \hat r_q unit vectors always point away from the charge. (For negative charges, the "-" sign that ends up in the electric field comes from the charge, not the unit vector). So if you're above the charge, \hat r_q points up. If you're below the charge, \hat r_q points down. In general, \hat r_q points radially away from the charge.

If your coursework doesn't use unit vector notation, don't worry about it then. Unit vectors are nothing more than a way to keep track of where things are pointing.

Regarding electric potentials:

Don't worry too much what I said earlier about needing calculus to use potentials. There are ways to teach and learn the concept of potentials without calculus.

I guess my point of all of that electric potential business was to say that when you get to the concept of electrical potential, greet it with a friendly smile. Electric potentials are here to help you.