Electric fields and electrostatic forces

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Homework Help Overview

The discussion revolves around calculating the magnitude of an unknown positive charge placed on the x-axis, which, in conjunction with a known negative charge, results in a zero electric field at a specified point. The subject area is electrostatics, focusing on electric fields and forces between charges.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between the charges and the distances involved in determining the electric field. There is an attempt to set up the equation for the electric field and to solve for the unknown charge. Some participants express uncertainty about the calculations and seek confirmation or further assistance.

Discussion Status

The discussion is active, with participants providing insights into the setup of the problem and the relevant equations. Some guidance has been offered regarding the conceptual understanding of the electric field, but there is no explicit consensus on the correctness of the calculations presented.

Contextual Notes

Participants note the importance of verifying calculations and assumptions regarding distances and charge magnitudes. There is an acknowledgment of the need for further responses to confirm the approach taken.

InfinitiMike069
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Electric fields and electrostatic forces (1 question)

I was wondering if someone can show me how to answer this problem. Thank you.


1)Two charges are placed on the x-axis, an unknown positive charge at x=0 cm and a negative 8.642e-6 C charge at 2.993 cm from the origin. Calculate the magnitude of the charge at x=0 cm which causes the total electric field at 11.917 cm to be zero.
 
Last edited:
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Show us what you did then I'm sure you'll get plenty of help!
 
Electric field,[tex]E = \frac{kq}{r^2}[/tex]
Since you have two particles with charges, let's call the unknown charge q1, and the known charge q2, where

Since E = 0, [tex]\frac{kq_1}{r^2} + \frac{kq_2}{r^2} = 0[/tex], where r is the distance from the charge to the point of zero electric field. Therefore...
For the first charge its .11917m away, and for the second charge its .11917m - .02993m = .08924m.

[tex]\frac{kq_1}{.11917^2} + \frac{k*-8.642e-6}{.08924^2} = 0[/tex]

The k's cancel, and therefore q1 = 1.54e-5 C.

I'm not 100% sure if I did this right, so wait for more responses before taking my word for it. :smile:
 
Last edited:
Conceptually it is correct although I haven't done the calculation so...
 

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