Troubleshooting: Finding the Equilibrium Position of a Charged Block on a Spring

[jnifong]

Here's the prob:

A 5.00kg block carrying a charge Q = 40.0 μC is attached to a spring for which k = 100 N/m. The block lies on a frictionless horizontal track, and the system in immursed in a uniform electric field of magnitude E = 5.00 x 10^5 V/m. a) The block is released from rest when the spring is unstretched (at x = 0). By what maximun amount does the spring expand? b) what is the equilibrium position of the block?

What I did:

a).

1 V/M = 1 N/C

F = qE (in electric field) = (40.0 x 10^-6 C)(5.0 x 10^5 N/C) = 20 N

F = -kx (Hooke's Law) so:

20N = -(100 N/M)x
x = 20N / -(100N/M)
x = -0.2m

But this is wrong! Where am I messing up.

b). I haven't attemped yet. How should I go about this one?

Thanks,

Jon

 

[Doc Al]

What you are trying to do for part a is actually the solution to part b! (Except for the sign.) In solving part a, you are making the same kind of error as was made in this post: https://www.physicsforums.com/showthread.php?t=77424 (See my comments there.)

 

[thepatientmental]

Hi Jon,

Thank you for sharing your work and questions. I can see that you have a good understanding of the concepts involved in this problem, but there are a few areas where you may have made some mistakes.

a) First of all, let's check the units. In your calculation, you used the value of 1 V/m = 1 N/C, which is correct. However, when you multiplied 40.0 x 10^-6 C by 5.0 x 10^5 N/C, you ended up with a unit of N^2/C instead of N. This is because you multiplied the electric field (N/C) by the charge (C), so the units should be N x C/C = N. Therefore, your value for the force should be 20 N.

Next, you correctly used Hooke's Law to equate the force from the electric field to the force from the spring. However, you mistakenly used a negative sign on the left side of the equation. Since the force from the electric field is acting in the same direction as the spring force, they should both be positive. Therefore, your equation should be:

20N = kx

Solving for x, we get:

x = 20N / 100 N/m = 0.2 m

So the maximum amount the spring expands is 0.2 m.

b) To find the equilibrium position of the block, we need to find the point where the forces from the electric field and the spring are balanced. This happens when the net force on the block is equal to zero. Using the same equation as before, we can solve for x:

0 = kx - qE

x = qE/k = (40.0 x 10^-6 C)(5.0 x 10^5 N/C) / 100 N/m = 0.2 m

So the equilibrium position of the block is also at 0.2 m.

I hope this helps to clarify where you may have gone wrong in your calculations. It's always a good idea to carefully check your units and signs when solving physics problems. Good luck with the rest of your assignments!