Strength and Direction of Electric Field

[TheCarl]

Homework Statement

What are the strength and direction of the electric field at the position indicated by the dot in the figure? Answer in component form.

Homework Equations

E=kQ/r^2

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) = 4.5E4

arctan(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.

 

[TSny]

Your work looks good to me. I don't see any mistake.

 

[SammyS]

Homework Statement

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[/PLAIN]

Homework Equations

E=kQ/r^2

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) = 4.5E4

arctan(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.

Round-off your final result to the correct sig. dig. if whoever/whatever is grading your work is anal about such things.

 

[TheCarl]

Turns out it was the comma in my vector notation. Once I removed the comma, the computer accepted it. Looks like I'm a bit rusty on that stuff. Thank you both for your help.