Sipko said:
Homework Statement
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 don't like them and they don't like me.
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 can not figure out the directions the magnitudes move in.
Homework Equations
r=1/2(√2 a) = a/√2
E=k*q/r2
ETotal = E1 + E2 + E3 + E4
The Attempt at a Solution
So far I only managed to do this:
r= 50cm/√2 = 3.55*10-2
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.
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 don't need to calculate the others.
So the next step involves knowing exactly which angles you need to calculate everything together. And that's what is giving me trouble.
[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.