Electric Energy and Spring Constant Problem

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Homework Help Overview

The problem involves two charged spheres mounted on identical springs on a frictionless table. Initially, the spheres are uncharged and spaced 0.0500 m apart. When charged with +1.60 microC, the spacing doubles. The objective is to determine the spring constant of the springs using energy equations related to kinetic, potential spring, and electric energies.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to apply energy conservation principles but expresses confusion over the mass and the electric potential energy (EPE) due to a lack of information provided in the problem. Some participants suggest focusing on the work done by the electric force and equating it to the potential energy in the springs.

Discussion Status

Participants are exploring different approaches to relate the electric force and potential energy to find the spring constant. Some guidance has been offered regarding the use of work-energy principles, but there is no consensus on a specific method due to varying levels of mathematical knowledge among participants.

Contextual Notes

There are constraints regarding the information provided in the problem, particularly the absence of mass and voltage values, which are complicating the attempts to find a solution. Additionally, the original poster notes that the chapter does not cover electric potential energy, suggesting a need for alternative methods.

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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