Calculating Electric Field Components at a Point: A College Physics Problem

[AshleyF708]

I'm having trouble getting started on a problem assigned to me in my College Physics class. First here's the problem:

Four charges occupy the four corners of a square(q1 is in the upperleft hand corner, q2 bottom left, q3 bottom right, q4 upper right). q1 = q3 = -q and q2 = q4 = +q , where q = 8 C. The sides of the square are all 2.2 meters long. Calculate the x and y electric field components at the point M. (M is between q2 and q3 on the line)

Ok now what I've come up with in the hours I've been staring at this problem:
What I have drawn is that there are three Electric fields coming off of this point M. One goes due east and is a combination of the electric field from q2 and q3. Another electric field goes Northwest towards q1. The last one goes Southwest in a line away from q4.

I have all these words and definitions in my head and I'm completely confusing everything to do with electric fields. Can someone please set me straight and help me through this problem? Hints, words of wisdom??

 

[himanshu121]

If u got struck Try doing this

Write each vector piece wise i.e the components along East & North

Apply the pythagoras to get the magnitude of the resultant and trigono for calculating the respectivr direction

 

[AshleyF708]

Ok, But first are my electric field vectors correct the way I have them set up? Should I figure out forces for each of those vectors?

 

[chroot]

Yes, Ashley, the directions are correct. Now, find the magnitude of each vector (use Coulomb's law). Finally, add the vectors together.

As himanshu says, it's easiest to add vectors when you select a basis and represent each vector with components in that basis. Adding the vectors then just involves adding the components.

- Warren

 

[AshleyF708]

Since M is the midpoint between the charges 2 and 3 I'm thinking the net electric field at M is zero because making rough digrams of the vectors they make a triangle. Am I on the right track or am I totally off base?

 

[himanshu121]

Yes if M is the Centre of Square

 

[AshleyF708]

Well M is the midpoint of the line between charges 2 and 3 not the center of the square

 

[chroot]

No, the total field at M is not zero.

Go ahead and represent each of the four vectors in the (east, north) basis and add them together.

- Warren

 

[himanshu121]

No its not 0 Can we have ur sol how u got 0

 

[AshleyF708]

Ok, how I see it looking at the electric field vectors coming from this point M, electric field vectors point away from positive charges and toward negative charges. SO From M there is a vector pointing east away from q2 and another one pointing east toward q3 so those two vectors can be easily added together because they both go east on the same line. So together that is one vector.
Now the q1 is negative so there is a vector from M going towards q1 making a second vector going northwest.
Last there is a vector going southwest away from q4 because it is positive.
The last two vectors when I draw them to add all the vectors i get a triangle.
SO yea I'm completely confused

 

[chroot]

Ashley,

Do what we continue to ask you to do. Express each vector as a pair of numbers.

The angle between the point M and the point q1 is 63.43 degrees, or (180 - 63.43) if you're counting from the east axis.

The angle between the point M and the point q4 is similarly 63.43 degrees, from the east axis.

First, verify these numbers. It's just basic geometry.

- Warren

 

[AshleyF708]

Ah forget it

 

[chroot]

We're trying to help you understand how to do this problem. Drawing the vectors, while useful for visualizing the problem, is not going to help you find the answer.

How did you teacher explain these problems?

- Warren

 

[AshleyF708]

That's the problem he hasn't explained how so I'm trying to teach myself.

 

[chroot]

Well, we're trying to help. Do you understand how to represent a vector with components?

- Warren

 

[AshleyF708]

Vaguely, it's been about two years since I took physics 1

 

[chroot]

Okay. The first step is to figure out the angle and direction of the each vector that sticks out of the point M. There are four vectors (although you can easily combine two of them, as you said).

First, list the angles (counted clockwise from the east direction) and magnitudes of each vector. Can you do this? It's just basic geometry, solving triangles.

- Warren

 

[AshleyF708]

ok I got 63.43 from the east

 

[chroot]

List the ANGLE and MAGNITUDE of EACH of the FOUR vectors.

- Warren

 

[AshleyF708]

Vector 1: 0 East
Vector 2: 0 East
Vector 3: 63.43 North of East
Vector 4: 63.43 South of East

 

[chroot]

Originally posted by AshleyF708
Vector 1: 0 East
Vector 2: 0 East
Vector 3: 63.43 North of East
Vector 4: 63.43 South of East

Don't use "North of East," since that will just confuse you. East is zero, north is 90, west is 180, and south is 270. Express the angle with just a single number, measured counterclockwise from east. I also don't think you meant "south of east."

And you did not list the magnitudes.

- Warren

 

[AshleyF708]

Vector 1: 0, 1.5 X 10^10 N/C
Vector 2: 0, 1.5 X 10^10 N/C
Vector 3: 63.43, 7.375 X 10^9 N/C
Vector 4: 116.57, 7.375 X 10^9 N/C

 

[chroot]

Getting there.

For the vector due to q2, I got the magnitude to be 5.94 * 10^10 N/C. Perhaps you made a calculation mistake?

For the vector due to q1 (which you call 4), I got the magnitude to be 1.18 * 10^10 N/C. Perhaps you made a calculation mistake?

The appropriate formule for finding the magnitude of the electric field is

[tex]|E| = \frac{Q}{4 \pi \epsilon_0 r^2}[/tex]

- Warren

 

[AshleyF708]

Vector 1: 0, 5.94 X 10^10 N/C
Vector 2: 0, 5.94 X 10^10 N/C
Vector 3: 63.43, 1.188 X 10^10 N/C
Vector 4: 116.57, 1.188 X 10^10 N/C

 

[chroot]

Almost there. What mistake did you make in calculating those magnitudes before?

Now, the vectors due to q2, q3, and q1 are all correct.

The one due to q4, however, appears to be backwards. Remember, it's positive, so its field points away from it. Thus, rather than pointing at 63.43 degrees, it should be pointing at (63.43 + 180) degrees.

Does this make sense?

Now, you have all four vectors' angles and magnitudes known. You can break the vectors down into components very easily.

A vector with magnitude m and direction d will have components (m cos(d), m sin(d)). You can verify this by drawing triangles.

In other words, three of the vectors are:

q2: (5.94 * 10^10, 0)
q3: (5.94 * 10^10, 0)
q1: (1.18 * 10^10 * -0.447, 1.18 * 10^10 * 0.894)

I'll let you do the one for q4.

You're almost done!

- Warren

 

[AshleyF708]

To get my magnitude before I was calculating force and then dividing by the charge which is where I got screwed up. I forgot about that equation you gave me because my prof hasn't covered it yet I only remembered it from reading.

But anywho, so the correct angle for q4 is 243.43 so I plug that into the coordinates and:
q4: (1.18 X 10^10 * -.447, 1.18 X 10^10 * -0.894)

I'm starting to see a light at the end of the tunnel.

 

[chroot]

You got it!

Now... to actually reach the light at the end of the tunnel, all you need to do is... add the numbers!

Add all the first (horizontal) components together, then add all the second (vertical) components together. The resulting vector is the total electric field at point M. If you want, you can convert the answer back into magnitude/direction format, but it should not be required.

- Warren

 

[chroot]

Originally posted by AshleyF708
To get my magnitude before I was calculating force and then dividing by the charge which is where I got screwed up. I forgot about that equation you gave me because my prof hasn't covered it yet I only remembered it from reading.

That's actually a perfectly valid way to do it -- consider that

[tex]E = \frac{F}{q} = \frac{Qq}{4 \pi \epsilon_0 r^2 q} = \frac{Q}{4 \pi \epsilon_0 r^2}[/tex]

The q just cancels out.

- Warren

 

[AshleyF708]

Ok so I get a total electric field (1.08 X 10^11 N/C, 0) Correct?

 

[chroot]

I agree.

And of course to convert this vector to magnitude/direction format is trivial. It points due east.

Do you understand how to do this sort of problem better now? The geometry can be a stumbling block for some people -- using sines and cosines in the right place and so on.

- Warren

 

[AshleyF708]

Yes Thank You, I suppose I am a little rusty on some of the basics. Plus, I tend to get a little confused setting up my problems. I was never really good at Physics. But Thanks a Million you were a great help!

 

[chroot]

Glad to hear. I'm glad you didn't give up!

- Warren

 

[himanshu121]

Im Glad That Chroot was there yesterday to help u.