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Strength and Direction of Electric Field

by TheCarl
Tags: electric, field, point charge, vector
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TheCarl
#1
Feb5-13, 06:44 PM
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) = 4.5E4
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.
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TSny
#2
Feb5-13, 07:12 PM
HW Helper
Thanks
P: 4,842
Your work looks good to me. I don't see any mistake.
SammyS
#3
Feb5-13, 08:26 PM
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PF Gold
P: 7,808
Quote Quote by TheCarl View Post
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) = 4.5E4
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
#4
Feb6-13, 08:04 PM
P: 21
Strength and Direction of Electric Field

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.


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