Electric Energy and Spring Constant Problem

In summary, two spheres mounted on identical horizontal springs and resting on a frictionless table have a spacing of 0.0500 m when uncharged. When each sphere has a charge of +1.60 microC, the spacing doubles. Assuming negligible diameter, the spring constant of the springs can be determined using the equation F=(kq_1q_2)/r^2, where K=(1/4piE_o) and E_o=8.85e-12. Through algebraic manipulation, the spring constant is calculated to be 920.8.
  • #1
imbroglio
5
0

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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  • #2
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.
 
  • #3
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 :)
 
  • #4
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
 

FAQ: Electric Energy and Spring Constant Problem

1. What is electric energy?

Electric energy is the energy that results from the flow of electric charge. It is a form of potential energy that can be converted into other forms of energy, such as light, heat, or mechanical energy.

2. How is electric energy measured?

Electric energy is measured in units of joules (J). One joule is equal to the energy transferred to an object when a force of one newton acts on that object over a distance of one meter in the direction of the force.

3. What is the spring constant?

The spring constant, also known as the force constant, is a measure of the stiffness of a spring. It is denoted by the letter k and is defined as the amount of force required to stretch or compress a spring by one unit of length.

4. How is the spring constant calculated?

The spring constant can be calculated by dividing the force applied to the spring by the amount of stretch or compression it undergoes. It can also be calculated using Hooke's Law, which states that the force exerted by a spring is directly proportional to the amount it is stretched or compressed.

5. How is electric energy related to the spring constant in a problem?

In a problem involving electric energy and spring constant, the electric energy is typically converted into mechanical energy stored in the spring. The amount of energy stored in the spring is determined by its spring constant, as a stiffer spring will store more energy for a given amount of stretch or compression. The relationship between these two factors can be described using the equation E = ½kx², where E is the energy, k is the spring constant, and x is the displacement of the spring.

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