What is the Zero Field Point Between Two Charged Particles on the x-axis?

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Homework Help Overview

The problem involves two charged particles located on the x-axis, where one particle has 34 excess electrons and is positioned at x = 100 μm, while the other has 17 excess electrons and is at the origin. The goal is to find the x-coordinate of a point between the two particles where the electric field strength is zero.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the placement of the zero field point and whether it lies between the charges or outside them. There are attempts to set up equations based on electric field strength and to analyze the behavior of these equations. Some participants question the signs of the charges and the implications for the electric field.

Discussion Status

There is an ongoing exploration of the correct setup for the equations governing the electric fields from both charges. Some participants have suggested different approaches to finding the zero field point, and there is recognition of the need to consider the signs of the charges in the equations. A potential solution has been identified, but there is still uncertainty regarding the calculations.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information they can use or the methods they can apply. There are discussions about the assumptions regarding the signs of the charges and the regions of interest along the x-axis.

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Homework Statement


Two particles are situated on the x axis. The particle q1, with 34 excess electrons, is situated at the point x = 100 μm. The other particle q2, with 17 excess electrons, is located at the origin. Give the x value of a point between the particles where the strength of the field they generate is zero.

Homework Equations


So kQ/r^2 is important here. The addition of electric field strength must give 0 N/C.



The Attempt at a Solution



I'm not even sure where that x value would fall into. Would it be between the two charges or to the right of -34e or to the left of -17e. kq1/x^2 +kq2/(x-100e-6)^2=0 is the equation I tried to solve for x, but the function is asymptotic to the x axis, so, the answer is infinity. But that wasn't the answer in the back of the book, so figure it out.
 
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Or I should've added x+100e-6 rather than subtracting it.
 
Pick an arbitrary point on the three regions of the line through the charges (to the left of, between, to the right of).

Without doing any numerics, sketch estimated field vectors for the contributions of both charges. For example, to the right of both charges, both field vectors will point in the same direction, to the left.

Determine the region where the two vectors have a chance of summing to zero, and concentrate your efforts there.
 
It will be between the charges. kq1/x^2 = -kq2/(x-100e-6)^2; where this holds true would be the x. But the calculator can't find the intersection.
 
Abelard said:
It will be between the charges. kq1/x^2 = -kq2/(x-100e-6)^2; where this holds true would be the x. But the calculator can't find the intersection.

Suppose you were to collect all your forces on one side of the equality and set to zero:

kq1/x^2 + kq2/(x-100e-6)^2 = 0

Both q1 and q2 have the same sign (they're both negative); Your expression can never be zero. One or the other of the terms needs to have its sign reversed.
 
OK, so kq1/x^2 = kq2/(100e-6-x)^2 would be a right equation. Now I got it. x=4.14e-5m. Thanks a lot.
 

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