Spring constant with point charges

[mujadeo]

Homework Statement

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

(a) Find the net electric field at point P, a distance d = 13 cm to the left of charge Qa.

This is no prob = -1364069.13N

(b) Find the force on Qb due to Qa .

No prob again = -1.837



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 = 9 cm.

(c) Calculate the spring constant ks of the spring.

this is what i can't figure out?

Homework Equations

The Attempt at a Solution

heres are the steps i did (its wrong though)

1. Spring constant is Fsp = -k delta s

2. i already know the force that the 2 charges exert on each other (= 1.837N)

3. delta s would be 21cm - 9cm = 12cm = .12m

4. then i just plugged in:
1.837 = -k(.12m)
k = -15.308

this is wrong.
please help

 

[Dick]

For a) the units of E field strength are not N. For b) that looks right, except you forgot to include the units, which are Newtons. For c) figure out the potential energy stored in the compressed spring and equate it to the change of potential energy of the moved charges. You can't assume that the force is a constant.

 

[mujadeo]

Dont I need the spring constant to get the PE stored in the spring though?
PE (in spring) = ( 1/2 k (delta s)^2)

 

[Dick]

The change in the PE of the charges gives you the change in the PE of the spring. That gives you the spring constant.