# Magnitude and Direction of an (E) Electric field of a square

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1. May 23, 2015

### Sipko

1. The problem statement, all variables and given/known data
What is E in Magnitude and Direction at the center of the square of (fig. 3-7). Assume that q = 10x10-8 C and a = 5 cm

Now I have am not well versed with vectors, I dont like them and they dont like me.
I can not figure out the directions the magnitudes move in.

2. Relevant equations
r=1/2(√2 a) = a/√2
E=k*q/r2
ETotal = E1 + E2 + E3 + E4

3. The attempt at a solution
So far I only managed to do this:
r= 5.0cm/√2 = 3.55*10-2
E-q=(8.99*109)*((-1.0*10-8 C)/(3.55*10-2 m)) = 7.13*104 N/C
E2q=(8.99*109)*(2*(-1.0*10-8 C)/(3.55*10-2 m)) = 1.43*105 N/C
And since the other magnitudes are the same I dont need to calculate the others.

So the next step involves knowing exactly which angles you need to calculate everything together. And thats what is giving me trouble.

Last edited: May 23, 2015
2. May 23, 2015

### collinsmark

Oh, please do make friends with the vectors. They are just waiting to shower you with love. All they ask is that you put some effort into getting to know them, and they will kindly return the favor with friendship and good grades.

I think you mean "5 cm" rather than "50 cm," but you seemed to have used the correct value in your calculation.

But there is another typo when you wrote the result down. The radius isn't quite what you specified; it's close, but not quite.

[Edit: there's a few typos in your intermediate steps such as using 1.0 x 10-8C instead of 10 x 10-8C, and forgetting to square the radius, but it seems you worked out some of these things correctly in your calculations (although you should still fix the slight radius calculation error discussed above).]

So yes, the next step is breaking up the individual electric field contributions into their respective x- and y-components.

In electrostatics, the electric field from a given, positive point charge points from the point charge in question to the test point (in this case the test point is point P). Don't forget to take the sign of the particular point charge into consideration. It's the opposite direction if the point charge is negative.

As far as the angles go, remember the charges lie on a square, and the test point P is at the center. Can you form any triangles between a given charge and the test point P? Do you see any right angles in your triangle? Remember, the sum of all angles in a triangle always sum to 180 degrees.

Also, there is some symmetry in the charge configuration. Taking note of this symmetry will cut the required calculation effort in half.

Last edited: May 23, 2015
3. May 23, 2015

### Sipko

Corrected 50cm to 5.0. Yes that was a typo.

For my professor it is more important to understand the problem rather than the exact number. So close should be close enough. :)
But for the sake of being exact here It should have been 3.53*10-2.

So and if we are going to Look at directions I imagine im looking at something like this:

Where positive charges have a somewhat "repelling" force negative ones have a somewhat "Pulling" force. (I know thats cheesy but I couldnt quite find the words in english)
And since all of these forces are directed whether outwards or towards a particular charge I can see that in my example the direction of the Electric field in this case is heading towards the "-X" direction. Correct so far?

4. May 23, 2015

### collinsmark

Yes, very nice. That is correct so far.

As you've determined, the y-components sum to zero due to symmetry.

So now, concentrate on the x-components of each vector. You'll still need to calculate those and sum them together.

5. May 23, 2015

### Sipko

So considering the vectors I found this:

I understand that the individual vectors of each charge seem to cancel out.
But I still don't quite grasp what I have to do to figure out the angles.
I know it's staring me in the face... :)

Last edited: May 23, 2015
6. May 23, 2015

### collinsmark

Triangles.

There are triangles in the diagram you made if you would prefer to use those. Or you can form new triangles too. Anyway, do you see any obviously 90 degree angles (i.e. right angles) in there anywhere?

Keeping in mind that all the angles of a triangle add up to 180 degrees. If one of the angles of the triangle is 90 degrees (opposite the hypotenuse), and the other sides of the triangle, besides of the hypotenuse, are of equal length (meaning the other two angles are equal to each other) what are those other two angles?

7. May 24, 2015

### Sipko

Of course. from point P to their respective charges form a triangle.
And the 45° Angle going from each charge towards the point P is quite obvious.

Edit: So does it mean that all of the angles are 45°?
So are all of them positive or are the angles from the -q charges towards the +2q charges negative and the angles from the +2q charges toward the -q charges positive?

Last edited: May 24, 2015
8. May 24, 2015

### Sipko

To be more exact about my edit above,
If I take the angle (between the x-vectors and the vectors heading towards the point P) how would I go about telling whether they are positive or negative?
And should they all be 45°?
Like:
ETotal= (-7.13*104)*(cos(-45°)+sin(-45°))+
+(1.43*105)*(cos(45°)+sin(45°))+
+....(1 more time each..)
Or simply the two above *2 would give the same result i suppose.
All that would equal to 4.4*105
this?

Last edited: May 24, 2015
9. May 24, 2015

### collinsmark

Before you start with the particular vector component, make sure you have the 2- or 3-dimensional vector in you mind, or better yet, drawn out on a diagram. Remember, for positive charges the vector starts from the point charge and ends at the test point; it's reverse for negative point charges.

Then to get the x-components of each vector, imagine that a light is cast on the vector such that the light rays are perpendicular to the x-axis. Thus imagine that the vector's shadow is on the x-axis. Then ask yourself whether the shadow points toward positive or negative x.

It turns out that in this particular problem, all x-components point in the same direction: along negative x. But that is just this particular problem.

Yes, in this problem, all the angles are 45°.

Something is not quite right with the math there.

For starters, recalculate the magnitudes. They seems to be off by about a factor of 2 or so. (sorry if I didn't make that clear in my earlier post.)

Then I'm not sure I follow the the "(cos(-45°)+sin(-45°))" parts, unless that is the result of factoring after adding two components together, in which case there wouldn't be a need for "+....(1 more time each..)" section.

For an individual vector component, there need only be a single trigonometric function (in this particular problem where the arrangement is in 2-dimensions).

10. May 25, 2015

### Sipko

Sadly I ran out of time and I had to send in what I had yesterday, otherwise all the other homeworks I have done already wouldn't have been accepted. I know Its bad to leave things undone but I have to say that this topic can be closed now. But thanks for your help anyway.