- #1

catie1981

- 45

- 0

**[SOLVED] more electric fields**

## Homework Statement

four particles form a square of edge length a = 5.20 cm and have charges q1 = 10.8 nC, q2 = -8.87 nC, q3 = 11.6 nC, and q4 = -10.4 nC. What is the magnitude of the net electric field produced by the particles at the square's center? q1 is in quadrant 2, q2 is in quadrant 1, q3 is in quadrant 4, and q4 is in quadrant 3

(SORRY THERE IS NO PICTURE, BUT HOPEFULLY THE DESCRIPTION WILL HELP YOU TO ENVISION THE FIGURE)

## Homework Equations

E=k(q/d^2) also multiplied by the sin or cos of a relevant angle

## The Attempt at a Solution

the value for d is obtained by the pythagoran theorem where d= sq.rt. 5.2^2+5.2^2= 3.68cm

I have

E2= 9e9(-8.87e-9/.0368^2) = -58948

E4= 9e9(-10.4e-9/.0368^2) = -69116

I sum those parts as the vector diagrm shows that the field for these two particles is pointing the same direction and get E-= -128064

E3= 9e9(10.8e-9/.0368^2) = 71775

E4= 9e9(11.6e-9/.0368^2) = 77091

Again summing the parts to get E+= 148866

I try to find the components by

E+(cos45)= 105264

E+(sin45)= 105264

E-(cos45)= -90555

E-(sin45)= -90555

then I use vector addition and obtain 14709i + 14709j

to find the magnitude of the field I used

sq.rt 14709^2 + 14709^2 = 20803 N/C (wrong according to the online program)

Where did I go wrong? The only example in our text uses symmetry to cancel out all but one charge, so when I look at it to compare why I'm getting the answer wrong, I can't complete my problem because I still have four charges to contend with, and don't know where I'm messing up. any help is great! Thanks!