Electric fields - determining k from known data.

[dmullin4]

Homework Statement

This is from a lab program for my class. Its set on a grid where I can place charges and use the mouse pointer to find the intensity of the field at any spot. I am using the E values I got from hovering my mouse over the indicated spot (x, y).

I have a positive 10 microcoulomb charge at (-140, 140). I have determined at (-100, 100), E=28.12 microcoulomb. They are asking me to find the constant of proportionality "k" at that point. I also have E=7.03 at (-60, 60), E=2.30 at (0, 0), and E=0.57 at (140, -140).

Homework Equations

E=[k*q*r(hat)]/r^2

The Attempt at a Solution

28.12 = [k*10*56.6]/56.6^2
k = 159.2

I'm not sure about my r(hat) variable. I understand it to be the vector pointing from the charge to the other point. I used the distance formula to get r=56.6. q=10 microcoulomb is given in the problem. When I performed this equation for the next data set (-60, 60), I got k=79.5. I thought k would be a constant that was the same for all equations, as in the force equation F=[k*q(1)*q(0)*r(hat)]/r^2 where k = 8.99x10^9.

Is k going to be constant, or will it change with different data?

and

How do I go about finding r(hat)?

and

What would be my units for k? (Nm^2/C^2)?

 

[Fightfish]

k should be a constant; check your calculations or program results again.
r(hat) is simply the unit vector indicating the direction of the electric field generated by the particle at that point - hence it has a magnitude of one and simply provides direction.
For the units of k, simply use dimensional analysis. (And, E, the electric field strength, is in Newtons per coulomb - force per unit (positive) charge, and not coulombs!)

 

[TwoTruths]

Yes, k is constant.

To find a unit vector in some direction, divide the vector by its magnitude. For example, take a vector directed along the x-axis <5,0>. Dividing by the magnitude gives you <1,0>, the unit vector directed along the x-axis.

Your units for k are correct.

In this problem, do you need to worry about the directional vector r-hat? What purpose does including r-hat in the formula serve?

 

[dmullin4]

Ok, I have recalculated with r(hat)=1 and got my k values to be very similar to each other. For (-100, 100), I got k=9008, for (-60, 60), I got 8992. Are my numbers off from the known value of k because of the microcoulombs? The known value is 8.99x10^6 and I am getting 8.99x10^3 so it makes sense that the micro would account for that.

I'm still not completely understanding the unit vector, r(hat), though. I remember working with the vectors while working with force and such, but they always confused me. If given the magnitudes I can add and multiply them and such, but figuring them out for myself is a bit of a mystery.


***To answer TwoTruths, I was given the formula with my lab assignment.

 

[TwoTruths]

A unit vector is basically a vector of magnitude 1 that says "go this way pl0x - I won't get in your way". Remember a force on a grid that was written like 5i-hat + 6j-hat? That says: "Please, go 5 units in the direction indicated by i-hat, then go 6 units in the direction indicated by j-hat! By the way, I won't augment your path at all, since my magnitude is 1. Thank you!" When you have the E = stuff * r-hat, you're saying: "The magnitude of E is kq/r^2, and r-hat tells me that it is directed in some direction."

Now then: can you tell me if r-hat matters for your purposes?

Have you done the dimensional analysis suggested by Fightfish? If so, you should know if something is wrong with your units. Remember that dimensional analysis looks like this:

1000 years * 2 decades/20 years = x decades

Include the units with the numbers in your calculations, and remember that k = Nm^2/C^2. Do your units check out? This is often a nice strategy for checking if your final equation for a homework/test problem is correct, by the way. Are my units right? Yes, maybe this is right. No? Dang, I screwed up somewhere. It can also point you in the direction of your mistake. (As in: woops, I found the volume [m^3] instead of the surface area [m^2], that's why I have this extra m left over!)