You have a lightweight spring whose unstretched length is 4.0 cm. First, you att

Question

You have a lightweight spring whose unstretched length is 4.0 cm. First, you attach one end of the spring to the ceiling and hang a 2.8 g mass from it. This stretches the spring to a length of 5.2 cm. You then attach two small plastic beads to the opposite ends of the spring, lay the spring on a frictionless table, and give each plastic bead the same charge. This stretches the spring to a length of 4.6 cm. What is the magnitude of the charge (in nC) on each bead? Express your answer using two significant figures.

Explanation / Answer

F = kx = ma = mg

F = mg = 2.8e-3*9.81 = 0.0275 N

Here, x = (5.2 - 4) cm = 1.2 cm = 0.012 m

After plug it,

F = kx

k = F/x = 0.0275/0.012 = 2.29 (spring constant)

Use that to find force of bead on either end,

F = kx

Here, x = (4.6 - 4) cm = 0.6 cm = 0.006 from bead on either end

And, k = 2.29 (spring constant)

F = 2.29*0.006 = 0.0137 N

q1 and q2 can be represented as q^2

Use equation F = Kq^2/r^2

K = 9*10^9 (not spring constant, a different K)

r^2= (.046)^2

Because that's the lenght it stretches

use F from above F = 0.0137 N

F = kq^2/r^2

q = sqrt(F/k)*r

q = sqrt(0.0137/9e9)*0.046 = 5.675*10^-8 C

q1 = q2 = 56.75 nC

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