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 1.8 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 4.8 cm . Required:
What is the magnitude of the charge (in nC) on each bead?

The magnitude of the charge is 54.9 nC.

Explanation:

The charge on each bead can be found using Coulomb's law:

Where:

q₁ and q₂ are the charges, q₁ = q₂  

r: is the distance of spring stretching = 4.8x10⁻² m

: is the electrostatic force

Now, we need to find . To do that we have that Fe is equal to the spring force ():

Where:

k is the spring constant

x is the distance of the spring = 4.8 - 4.0 = 0.8 cm

The spring constant can be found by equaling the sping force and the weight force:

where x is 5.2 - 4.0 = 1.2 cm, m = 1.8 g and g = 9.81 m/s²

Now, we can find the electrostatic force:

And with the magnitude of the electrostatic force we can find the charge:

Therefore, the magnitude of the charge is 54.9 nC.

I hope it helps you!  

a charge 3q is at the origin, and a charge −2q is on the positive x axis at x=a. part a where on the x-axis would you place a third charge so it would experience no net electric force? enter an expression for the exact answer, which will involve a square-root sign. (do not enter an approximate decimal answer.)

answer: 2×10^-2kg

explanation: density= mass per volume

vol=mass/density

converting to s.i units we have;

3÷1000kg = 0.003

1.5÷100m = 0.15

0.003/0.15 = 0.02kg