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
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
To solve this problem, we need to consider both the electrostatic force between the charges and the force exerted by the spring.
Let's break down the steps to find the correct answer:
Step 1: Calculate the electrostatic force between the charges.
The electrostatic force between two charges can be calculated using Coulomb's law:
F = (k * Q1 * Q2) / r^2
where F is the force, k is the electrostatic constant (9 * 10^9 Nm^2/C^2), Q1 and Q2 are the charges, and r is the distance between the charges.
In this case, the charges are both 34 * 10^-6 C, and the distance between them is the equilibrium length of the spring (0.5 m). So, the electrostatic force is:
F_electrostatic = (9 * 10^9 * (34 * 10^-6)^2) / (0.5)^2
Step 2: Calculate the tension in the spring.
The tension in the spring can be determined using Hooke's Law:
F_spring = k * x
where F_spring is the force exerted by the spring, k is the spring constant (150 N/m), and x is the displacement from equilibrium.
Step 3: Set up equilibrium condition.
For the system to be in equilibrium, the electrostatic force and the spring force should be equal:
F_electrostatic = F_spring
Step 4: Solve for x.
Substitute the expressions for the electrostatic and spring forces into the equilibrium condition equation:
(9 * 10^9 * (34 * 10^-6)^2) / (0.5)^2 = 150 * x
Simplify the equation and solve for x.
By following these steps, you should be able to arrive at the correct answer.