Strength and direction of an electric field at a particular point problem

  • #1
Hello all - this is my first time giving this forum a try. I've been using another site for a while but it's just too spotty and not enough folks there care to give decent responses, so here I am :)

Homework Statement



http://img684.imageshack.us/i/27p28.jpg/

What are the strength and direction of the electric field at the position indicated by the dot in the figure?

a) Give your answer in component form. (Assume that x-axis is horizontal and points to the right, and y-axis points upward.) Express your answer in terms of the unit vectors i and j.

b) Give your answer as a magnitude and angle measured cw [Edit: ClockWise?] from the positive x-axis.

c) Give theta, the degrees clockwise from the positive x axis.


Homework Equations



E_1x = E_1*cos(theta) = k*(q1 / r1^2) * cos(theta)

At least, I believe this is the main equation you need...?

The Attempt at a Solution



The book I'm using is confusing about this. I've tried the idea of calculating the fields for each of the points (the two positive charges and the electric charge), but as far as the angles go and how the negative charge will pull away from the rest of those charges...I start to get a headache. How does that factor in with the rest of the angles? Does it, perhaps, pull away straight up and leave the x components of the other one to push it out? Why am I not getting the right answer for that then when I try it, in such a case... Does anyone have an idea?
 

Answers and Replies

  • #2
ideasrule
Homework Helper
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The book I'm using is confusing about this. I've tried the idea of calculating the fields for each of the points (the two positive charges and the electric charge), but as far as the angles go and how the negative charge will pull away from the rest of those charges...I start to get a headache. How does that factor in with the rest of the angles? Does it, perhaps, pull away straight up and leave the x components of the other one to push it out?

Yes, the electric field of the negative charge points up at P. However, you don't need to worry about this. Since Ex=kq/r^2 * cos(theta), you can simply insert the negative charge into q and get the right answer. There's nothing special in the equation that makes it only work for positive charges.

Why am I not getting the right answer for that then when I try it, in such a case... Does anyone have an idea?

Can you show all your work? It might just be an algebra mistake.
 
  • #3
Well I actually just think I might be on the right track for these parts at least...though honestly I'm really quite confused still, having not done anything like this in years (you'll see that from my work, I'm sure...). My work:

<Note: I've labeled the charges as such: E1 is for the field from the lower left charge, E2 from the upper left, and E3 from the upper right>

E_netx = E_1x + E_2x

E_1x = 9*10^9 * 5*10^-9 / 0.02^2 = 112500 N/C

E_2x = 9*10^9 * 10*10^-9 * cos(63.4degrees) / 0.045^2 = 1990 N/C

Therefore E_netx = 114490 N/C


E_nety = E_2y + E_3y

E_2y = 9*10^9 * 10*10^-9 * cos(26.6) / 0.045^2 = 3974 N/C

E_3y = 9*10^9 * (-5*10^-9) * cos(0) / 0.04^2 = -28125

Therefore E_nety = 3974 - 28125 = -24151 N/C

....and honestly I have no idea what to do with this information o.o I'd need to find the overall angle as to where the direction of that field would go, and for some reason I'm just not getting that. How horribly wrong am I, and what can I do to correct myself and push forth?
 

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