Net Electric Field: Find Magnitude @ x=6cm

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Homework Help Overview

The problem involves calculating the net electric field at a specific point on the x-axis due to two point charges, one positive and one negative, placed at defined positions. The context is rooted in electrostatics and vector addition of electric fields.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to apply a formula for the electric field but indicates that their solution is incorrect. Some participants question the accuracy of the vector addition and suggest reviewing the calculations.

Discussion Status

Participants are actively engaging in checking the calculations and discussing the importance of sign conventions in vector addition. One participant has indicated they resolved their issue by adjusting the signs in their equation, suggesting progress in understanding the problem.

Contextual Notes

There is an emphasis on the importance of correctly setting up the problem and ensuring that all vector components are accurately accounted for, particularly regarding signs in the calculations.

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Two charges are placed on the x axis. One charge (q1=+8.45 mC) is at x1=+2.55 cm and the other (q2=-21.4 mC) is at x2=+9.01 cm. Find the magnitude of the net electric field at x=6 cm.

I used this equation:
(x is in cm)

(N/C) = {(k * 8.45*10^-6) / [(x+2.55)^2 / 100]} - {(k * 21.4*10^-6) / [(x-9.01)^2 / 100]}

and my solution is not correct.
 
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Did u add the 2 vectors correctly??Post the first part of your calculations.It's essential.

Daniel.
 
figured it out, thx.
i just needed to change
some signs in that equation.
 
Well,u see,that's why it was important to check your initial setting...Some or maybe one sign wrong can make a whole lot of difference,and with vectors u always have to keep an eye out...

Daniel.
 

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