Question

Two 34 −ìC charges are attached to the opposite ends of a spring of spring constant 150 N/m and equilibrium length 50 cm.

that is the question I should probably know this but my brain is fried so far all I have is

Original length = 50 cm = 0.5 m
Let the spring stretch by distance x.
Then new length = 0.5 + x
Each charge Q = 34 * 10^-6 C
Spring constant k = 150 N/m

Electrostatic force = 9*10^9*Q^2/(0.5 + x)^2
Tension in the spring = (1/2)k*x^2
For equilibrium,
9*10^9*Q^2/(0.5 + x)^2 = (1/2)k*x^2

9*10^9* (34*10^-6)^2/(0.5 + x)^2 = (1/2)*150*x^2
10.404/(0.5+x)^2 = 75 * x^2
10.404/75 = x^2 * (0.5+x)^2
Taking square root on both sides,
sqrt(10.404/75) = x * (0.5 + x)
0.372 = x * (0.5 + x)
0.372 = 0.5 x + x^2
x^2 + 0.5 x - 0.372 = 0
x = [-0.5 +- sqrt(0.5^2 + 4*0.372)]/2
x cannot be negative.
x = [-0.5 + sqrt(1.738)]/2
x = (-0.5 + 1.32)/2
x = 0.82/2
x = 0.41 m
x = 41 cm
Ans: 41 cm

But it is not correct Im losing my mind over this stuff any help would be great

Answer 1

Tension in the spring = (1/2)k*x^2
For equilibrium,
9*10^9*Q^2/(0.5 + x)^2 = (1/2)k*x^2
that is your error. Tension in the spsring is kx and for equilibrium,
kx=9*10^9*Q^2/(0.5 + x)^2