- #1
mr_coffee
- 1,629
- 1
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 field 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 field 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?
Thanks.
^ 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 field 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?
Thanks.