Coulomb vs Hooke
There are two identical charges attached to the ends of a spring. How far apart are they if the system is in equilibrium (neglecting gravity)?
System Parameters (All in SI units):
| Charge | Coulomb's constant | Spring constant | Spring unstretched length |
| q = 1 0 − 4 C | k = 9 × 1 0 9 Nm 2 C − 2 | α = 9 0 Nm − 1 | L = 1 m |
The answer is 1.4656.
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1 solution
But why you have not taken the distance b/w the charges initially.i.e 1m?
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Strictly speaking, one doesn't need to because only the equilibrium position matters. However, you could very well start from the unstretched condition and then allow the system to undergo damped oscillation until the final state is reached. Some "viscosity" would have to be included to keep it from oscillating forever. Perhaps you could run that simulation and post your result. The steady-state value should be the same. If not, let me know and I will post it. Interesting idea. Thanks.
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I have updated my main solution with the new simulation results.
thanks for your new solution
i got it (what you tried to explain)
Not to be a noob, but how did you solve that final cubic expression?
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Newton's method. Formal solution of cubic equations is non-trivial to say the least.
Missed because of calculation mistakes.
Electric force = Spring force.
Let x be the elongation, then,
(
1
+
x
)
2
9
∗
1
0
−
9
∗
(
1
0
−
4
)
2
=
x
∗
9
0
.
⟹
(
1
+
x
)
2
1
=
x
.
S
o
l
v
i
n
g
t
h
e
C
u
b
i
c
,
x
=
.
4
6
5
6
.
So the distance between the charges is 1.4656.
Also, just for fun, here's a small time step simulation for this problem. One charge is fixed at the origin and the other can oscillate, subject to some-artificially induced damping. The simulated separation between the charges settles down to the hand-calculated equilibrium value.