Physics Question Related to Charges and a Spring?

Question

I keep trying to figure out this problem but I can't get the right answer... I think I'm doing it right :/

Two charges Qa = 1 µC and Qb = -3 µC are placed on the x-axis with a separation of a = 21 cm.

The charges Qa and Qb are now attached to the ends of a spring whose unstretched length is s0 = 21 cm. With the charges attached, the spring compresses to an equilibrium length s1 = 10 cm.

Calculate the spring constant ks of the spring.

Okay. So I tried using the potential energy for the two charges and set that equal to the potential of the spring:

(9e9)(1e-6)(3e-6)/(.10) - (9e9)(1e-6)(3e-6)/(.21) = .5 * k * (.21 - .10)^2

I keep getting k= 23.3776 Nm, but it's wrong.

If someone could let me know where I'm going wrong, that'd be great. Thanks!

Answer 1

Note: express k's units as N/m, not Nm

r1=0.21m

r2=0.10m

let k=9*10^-3

From conservation of energy:

k*(q1*q2)/r1-k*(q1*q2)/r2 = (1/2)*ks*x^2

The potential energy of the charges decreased by

(9*10^-3)*(1*-3)/(0.21)-(9*10^-3)*

(1*-3)/(0.10) = 0.141428571428571

Solving for ks

(1/2)*ks*x^2=0.141428571428571

ks=(0.141428571428571*2)/(0.21-0.10)^2

= 23.3766233766233 N/m ***

check:

The potential energy of the spring increased by

(1/2)*ks*x^2

(1/2)*23.3766233766233*(0.21-0.10)^2

= 0.141428571428571

check: using equilibrium of forces:

Force between charges at 0.10 m

(9*10^-3*1*-3)/(0.10)^2 = -2.7 N

=

force on spring:

-2.7=-ks*x

solve for ks

ks=2.7/(0.21-0.10) = 24.54 N/M ********

Answer should be correct using Conservation of energy

or balance of force under ideal conditions:

Answer 2

ok its been a while but this is how i looked at it.  when the spring is compressed, the spring force equals the coulomb force ( i.e. kx = Qa*Qb / (4*pi*e_0 *r^2) ).   so you solve for k.  r is 0.1m  x is the difference (s0 - s1). i get (1e -6) ( 3e -6)(9e 9)/[(.11)(.1)(.1)] = 24.55 N/m.

So the moral of the story is equate forces, not energy.  if this answer is wrong let me know so i can help you find the right one.