# Strength and Direction of Electric Field

by TheCarl
Tags: electric, field, point charge, vector
 P: 21 1. The problem statement, all variables and given/known data What are the strength and direction of the electric field at the position indicated by the dot in the figure? Answer in component form. 2. Relevant equations E=kQ/r^2 3. The attempt at a solution E(+5) = (8.99E9)(5E-9)/(0.02^2) = (1.1E5i, 0j) E(-5) = (8.99E9)(5E-9)/(0.02^2) = (0i, 2.8E4j) r(+10) = SQR(0.02^2+0.04^2) = 0.0447 E(+10) = (8.99E9)(10E-9)/(0.0447^2) = 4.5E4arctan(2/4)=26.57DEG = (4.5E4sin(26.57)i, 4.5cos(26.57)j) = (2.0E4i, -4.0E4j)E(+5) + E(-5) + E(10) = (1.3E5i, -1.2E4j) Mastering Physics is telling me that this answer is wrong and I want to know where I screwed up.
 HW Helper Thanks PF Gold P: 3,792 Your work looks good to me. I don't see any mistake.
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 Quote by TheCarl 1. The problem statement, all variables and given/known data What are the strength and direction of the electric field at the position indicated by the dot in the figure? Answer in component form. [ IMG]http://session.masteringphysics.com/problemAsset/1384377/2/27.P28.jpg[/IMG] 2. Relevant equations E=kQ/r^2 3. The attempt at a solution E(+5) = (8.99E9)(5E-9)/(0.02^2) = (1.1E5i, 0j) E(-5) = (8.99E9)(5E-9)/(0.02^2) = (0i, 2.8E4j) r(+10) = SQR(0.02^2+0.04^2) = 0.0447 E(+10) = (8.99E9)(10E-9)/(0.0447^2) = 4.5E4arctan(2/4)=26.57DEG = (4.5E4sin(26.57)i, 4.5cos(26.57)j) = (2.0E4i, -4.0E4j)E(+5) + E(-5) + E(10) = (1.3E5i, -1.2E4j) Mastering Physics is telling me that this answer is wrong and I want to know where I screwed up.
The problem may well be too much round-off error.

Keep at least two extra digits for all intermediate results.