Field and force with three charges

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SUMMARY

The discussion focuses on calculating the electric field at the location of charge q1 (3 nC) due to charge q2 (7 nC) using Coulomb's Law. The correct formula is E = 9e9 * (q/r^2) * r-hat, where the distance must be converted from centimeters to meters. The user initially miscalculated the distance, leading to an incorrect electric field value. The clarification provided by another user emphasized the importance of proper unit conversion from centimeters to meters.

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  • Coulomb's Law for electric fields
  • Understanding of electric charge units (nC)
  • Vector representation of forces and fields
  • Unit conversion between centimeters and meters
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  • Learn about electric field superposition principles
  • Study vector components in physics for better understanding of directionality
  • Practice unit conversions in physics problems, especially for electric fields
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Homework Statement



At a particular moment, three small charged balls, one negative and two positive,are located as shown in the figure. q1 = 3 nC, q2 = 7 nC, and q3 = -5 nC.

What is the electric field at the location of Q1, due to Q2?

y
q1
|
|
|
q2______q3____x

Okay that is the given diagram the if q2 is at (0,0) then q1 is at (0,4); q3 at (3,0);
and the distances are in centimeters

Homework Equations



I'm using Coulomb's Law which states that E = 9e9*(q/r^2)r-hat

The Attempt at a Solution



So I thought all I have to do is plug in the values

[9e9*(7e-9/(.004)^2)]*<0,4,0> = <0,3937500,0>N/C

please help me, I've been staring at this for a while and I cannot figure out why I don't get the right answer, I shouldn't have to consider the other charges because they don't affect the field created by q2 just the net field, so why is this not correct I have no idea.

Thanks Again for all the help guys
 
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In what units are the components of the position?

ehild
 
The components of the positions are in centimeters. (cm)
 
If so, 4 cm = 0.04 m. And you have to use the unit vector in the direction of <0,4,0>.

ehild
 
Oh yes your right, I just kept converting the cm to m wrong, putting an extra zero in front of it like they were mm or something. Thank You so much it really helps having someone else look at this. I knew I had to be making some stupid mistake.
 

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