Electric Energy and Spring Constant Problem

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SUMMARY

The discussion centers on calculating the spring constant of two charged spheres mounted on identical springs. When uncharged, the spheres are 0.0500 m apart, but this distance doubles when each sphere carries a charge of +1.60 microC. The relevant equations include potential energy in springs (PE_sp = (1/2)kx^2) and electric potential energy (PE_e = k(q1)(q2)/r^2). The final calculation yields a spring constant of approximately 920.8 N/m.

PREREQUISITES
  • Understanding of electric potential energy (EPE) and its relation to charge and voltage.
  • Familiarity with Hooke's Law and spring potential energy equations.
  • Basic knowledge of forces and Newton's laws of motion.
  • Ability to manipulate algebraic equations and solve for unknowns.
NEXT STEPS
  • Study the derivation and application of Hooke's Law in various physical systems.
  • Learn about electric forces and fields, particularly Coulomb's Law.
  • Explore the concept of electric potential energy and its calculations in electrostatics.
  • Investigate the relationship between force, mass, and acceleration in dynamic systems.
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone involved in mechanics or electrostatics, particularly those tackling problems involving spring constants and electric forces.

imbroglio
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Homework Statement


Two spheres are mounted on identical horizontal springs and rest on a frcionless table, as in the drawing. When the spheres are uncharged, the spacing between them is 0.0500 m, and the springs are unrestrained. When each sphere has a charge of +1.60 microC, the spacing doubles. Assuming that the spheres have a negligible diameter, determine the spring constant of the springs.

Homework Equations



E=KE_t+KE_r+U_g+PE_sp+PE_e

Energy is equal to the sum of all energies, translational kinetic, rotational kinetic, gravitational potentional, potential spring, and potential electric.

Potential gravitational and Rotaional Kinetic are not applicable

KE_t=(1/2mv^2)
PE_sp=(1/2)kx^2
PE_e=[k(q_1)(q_2)]/r^2

The Attempt at a Solution



E=KE_t+PE_sp+PE_e
Ef=[(1/2)m_1f V_1f^2 +(1/2)M_2f V_2f^2]+[1/2kx_f^2]+[(k_e)(q^2)]/r_f^2]

Ei=0+[1/2kx_i^2]+[(k_e)(q^2)]/r_i^2]

Ef=Ei
[(1/2)m_1f V_1f^2 +(1/2)M_2f V_2f^2]+[1/2kx_f^2]+[(k_e)(q^2)]/r-f^2]=[1/2kx_i^2]+[(k_e)(q^2)]/ri^2]

(1/2)[m_1f V_1f^2 +M_2f V_2f^2]+[kx_f^2]+[(2k_e)(q^2)]/rf^2]= (1/2)[Kx_i^2]+[(2k_e)(q^2)]/r_i^2]

(m1=m2)

[mf Vf^2]+[kx_f^2]+[(2k_e)(q^2)]/rf^2]=[(2k_e)(q^2)]/r_i^2]

Mv^2+x_f^2+(1/r_f^2)=(1/r_i^2)

So then...

(1/2) [m_1v_1f^2+m_2v_2f^2]+Kx_f^2+[(k_e)(q^2)]/r^2]=0

K=-[[k_e(q^2)/r^2]-[mv_f^2]]/x^2

K=-[(8.99e9)(2.56e6)]/(.01)]-m(1.73^2)

k=-29.929(x+7.689)

and I'm stuck...I don't know what I did wrong really...I don't know how to find the mass though...

Thanks!
 
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Consider only one of the spheres. When the electric force pushes on it it does a certain amount of work, right? You could figure that out through integration (or if you don't know any calculus, I'm sure there is some formula you could use). Set that equal to the increase in potential energy in one of the springs and youve got your answer.
 
That sounds like a fantastic idea. However, with my lack of knowledge in calculus and the limitations of algebra, I cannot find the EPE. It is not given in the problem. In algebra we only know EPE to be EPE=qV...the voltage is not given in the problem.

In addition, this chapter of which the homework is assigned, does not cover EPE. Thus, there must be an alternative solution...any ideas?

Thanks for your help though, I do appreciate it :)
 
My friend from school enlightened me on this matter and it is properly performed with algebraic physics as follows.

F=(kq_1q_2)/r^2

K=(1/4piE_o) when E_0=8.85e-12

F=[(K(1.6e-6)^20/(0.1^2)]
F=2.302

F=k_spx^2
K=(f/x^2)
K=(2.302/0.0025)
K=920.8

yay
 

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