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catie1981
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[SOLVED] more electric fields
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)
E=k(q/d^2) also multiplied by the sin or cos of a relevant angle
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!
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!