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^ 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?

Thanks.