Electric Field Produced by a Ring

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SUMMARY

The discussion focuses on calculating the electric field produced by a uniform circular ring of charge, specifically with a charge of Q=5.60 microCoulombs and a radius of R=1.30 cm. The electric field equation is given as Ez = (kQz)/(R^2 + z^2)^(3/2). The user attempts to simplify this for cases where z is much smaller than R, leading to the expression Ez = (-kQz)/(R^3). The correct constant A is derived as A = (kQ)/(R^3), resulting in a numerical value of 2.3 x 10^10 kg/s². The user is reminded to consider the charge of the electron when calculating the force.

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  • Understanding of electric fields and forces
  • Familiarity with Coulomb's law and constants
  • Basic calculus for limits and simplifications
  • Knowledge of units for charge and force
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  • Learn about the implications of charge polarity on force calculations
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[SOLVED] Electric Field Produced by a Ring

1. Homework Statement

A uniform circular ring of charge Q=5.60microCoulombs and radius R=1.30 cm is located in the x-y plane, centered on the origin as shown in the figure.

If z is much smaller than R then E is proportional to z. (You should verify this by taking the limit of your expression for E for z much smaller than R.) If you place an electron on the z-axis near the origin it experiences a force Fz=-Az, where A is a constant. Obtain a numerical value for A.


2. Homework Equations

Ez= (kQz)/(R^2+z^2)^(3/2)


3. The Attempt at a Solution

Since R is much greater than z and the electron carries a minus sign, Ez simplifies to
(-kQz)/(R^3)

Then Eqauting Fz and Ez gives (-kQz)/(R^3)=(-)Az which then simplifies to
(kQ)/(R^3)=A

Plugging in the numbers gives (9*10^9*5.6*10^-6 C)/(.013 m)^3= 2.3*10^10 kg/s^2

What am I doing incorrectly? Any help is appreciated.
 
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Well, for one, you haven't taken into account the charge of the electron - it's magnitude to be precise. The question asks you to find the force F = qE, where q is the charge of the test particle.

Secondly, you don't have the right unit for force.
 
Thanks for the help.
 

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