Charged particle suspended from elastic cord

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SUMMARY

A charged particle with a mass of 200g and a charge of 25 µC is suspended from an elastic cord with a force constant of 5 N/m in a vertical electric field of 3.7 x 10^5 N/C. The equilibrium condition is established by balancing the forces acting on the particle: the gravitational force (mg) acting downward, the restoring force from the elastic cord (kx) acting upward, and the electric force (F_e) also acting upward. The equation F_e + kx = mg is used to solve for the extension (x) of the cord when the particle is in equilibrium.

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  • Familiarity with Hooke's Law
  • Basic knowledge of electric fields and forces
  • Ability to solve algebraic equations
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  • Learn about electric field strength and its effects on charged particles
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Homework Statement



A charged particle with mass m = 200g and charge q = 25 µC is suspended from an elastic cord with force constant K = 5 N/m in a vertically upward electric field E of magnitude 3.7 x 10^5 N/C. What is extension of the cord when the particle is in equilibrium?

Homework Equations





The Attempt at a Solution

 
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Hi

first draw a free body diagram. there are three forces acting on the body. [itex]mg[/itex] downwards , the force of restoration due to the elastic cord , according to Hook's law is
[itex]kx[/itex] upward and the force due to electric field [itex]F_e[/itex] upward. when in equilibrium, they balance.

[tex]F_e+kx = mg[/tex]

solve for x, the extension.
 

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