Electric Field and Force of a Charged Particle Near a Stationary Ring

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SUMMARY

The discussion focuses on the electric field and force experienced by a negatively charged particle located at the center of a stationary positively charged ring. The electric field E along the axis of the ring is derived as E = kQx/(a^2 + x^2)^(3/2), and for small displacements (x << a), it simplifies to E ≈ kQx/a^3, confirming that E is proportional to x. The force on the particle is calculated as F = -qE, leading to F = -q(kQx/a^3). The conditions for simple harmonic motion (SHM) are established by showing that the acceleration is proportional to the displacement, with the angular frequency w defined as w = sqrt(qkQ/a^3).

PREREQUISITES
  • Understanding of electric fields and forces in electrostatics
  • Familiarity with the concept of simple harmonic motion (SHM)
  • Knowledge of the mathematical representation of forces and fields (e.g., Coulomb's law)
  • Basic calculus for integration and differentiation
NEXT STEPS
  • Study the derivation of electric fields for different charge distributions, specifically charged rings
  • Learn about the conditions for simple harmonic motion and its mathematical formulation
  • Explore the concept of potential energy in electric fields and its relation to forces
  • Investigate the effects of varying charge distributions on the stability of charged particles
USEFUL FOR

Physics students, electrical engineers, and anyone interested in electrostatics and dynamics of charged particles in electric fields.

bodensee9
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Hello:

I have the following. A stationary ring of radius a lies in the yz plane and has a uniform positive charge Q. A small particle that has mass m and a negative charge -q is located at the center of the ring. (a) show that is x << a the electric field along the axis of the ring is porportional to x. (b) find the force on the particle as a function of x. (c) show that if the particle is given a small displacement in the +x direction, it will perform SHM.

So for (a), do I do, since the E for a ring is k*Q*x/(a^2+x^2)^(3/2), where x is the displacement on the z axis, and a is the radius. So that's a because if x is very small then the equation is basically k*Q*x/a^3. So this is porportional to x.

(b) Then wouldn't the force just be q*E, and since there's a negative charge -q here, wouldn't the F = q*k*Q*x/a^3?

(c) So to show SHM, I need to show that acceleration = some constant w^2*displacement. So couldn't I just set q*k*Q/a^3 as w?

Thanks!
 
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Your answer for (a) is correct provided you got the expression for E correct (I didn't check it). (b) is odd since x=0 and it is evident by symmetry that that E-field at the centre of the ring where the particle is located is clearly 0. Unless they mean (if the particle was instead placed at some x=b), then you would have to perform the integration to get the value.

(c) is nearly correct. Just note you have to take the mass of particle into consideration. Be careful of the missing square.
 
THanks!
 

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