Determining the electric field at point a and b, stuck on finding resultant E!

  • Thread starter mr_coffee
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  • #1
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Hello everyone, i'm having problems finding The resultant intensity at point b, i know its the vector sum of E1 and E2. q1 = -6x10^-9; q2 = 6x10^9 they are 12cm apaprt. Determine the electrict feild at A and B, i already figured out A. This is the free body diagram.


^ E2
..\
...\
...(B)
....|
....|
....v E1





(q1)----------------------------------(q2)


The angle that E2 forms is a 37 degree angle.
E1 is 9cm from q1.
Point B is 15 cm from q2 diagnoally.

Okay I figued the feild intensity at B due to q1 is directed downward and is equal to
E1 = kq1/r^2 = -[(9x10^9)(6x10^-9)]/(9x10^-2m)^2;
E1 = -.667x10^4 N/C;

To find E2
E2 = [(9x10^9)(6x10^-9)]/(15x10^-2m)^2;
E2 = .240x10^4N/C 37 degree N of W.


Now this is where I get lost...
I need to find the X and Y component of E2 which seems easy enough....but i f it up anyways.

So i figure, you have a triangle that looks like this:

^
|E2y
| 37 degrees
<------
E2x

E2y = E2*sin37;
E2x = E2*cos37;

E2x = -.192x10^4 N/C
E2y = .144x10^4 N/C

They did the following:

Ex = -E2x = -(.240x10^4 N/C)*cos37;
Ex = -.144x10^4 N/C; //is this a misprint? :bugeye:

Thanks.
 

Answers and Replies

  • #2
Doc Al
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mr_coffee said:
They did the following:

Ex = -E2x = -(.240x10^4 N/C)*cos37;
Ex = -.144x10^4 N/C; //is this a misprint? :bugeye:
Looks like a misprint to me. It's obviously wrong. (.240x10^4)*cos37 does not equal .144x10^4 !

What book are you using?
 
  • #3
GCT
Science Advisor
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just aside, in the future you might want to use a scanner to illustrate the diagrams by posting it as an attachment.
 
  • #4
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GCT said:
just aside, in the future you might want to use a scanner to illustrate the diagrams by posting it as an attachment.
hah good thinking! The book is called Applied physics, it was published in 1979 :uhh: I'm trying to polish up in physics so i don't screw up physics 212 when i go back in the fall.

I also had another question about the problem....
If the problem only gives the distance from the 2 charges, 12cm...how did they figure out:

#1. The angle 37 degrees
#2. The distance from point B to charge 2?
 
  • #5
Doc Al
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mr_coffee said:
I also had another question about the problem....
If the problem only gives the distance from the 2 charges, 12cm...how did they figure out:

#1. The angle 37 degrees
#2. The distance from point B to charge 2?
If all they gave was the distance between the two charges, then you couldn't figure out the angle. But they also give you the fact (I assume) that point B is 9cm above q1. Thus you have two sides of a right triangle: the bottom is 12, the left side is 9. You have all you need to figure out the angle and the hypotenuse of that triangle (which is the distance from B to q2) using a little trig (or the Pythagorean theorem).

This particular right triangle is a popular one: The sides are in the ratio of 3-4-5, thus it's often called the "3-4-5 right triangle". :smile:
 
  • #6
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Thanks for the info! I really gota work on my trig!
 
  • #7
GCT
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Thanks...work on my trig!
....and vectors
 

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