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 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?

Answer 1

Answer:

The magnitude of the charge is 54.9 nC.

Explanation:

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

$F_{e} = \frac{k*q_{1}q_{2}}{r^{2}}$

Where:

q₁ and q₂ are the charges, q₁ = q₂  

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

$F_{e}$: is the electrostatic force

$F_{e} = \frac{k*q^{2}}{r^{2}} \rightarrow q = \sqrt{\frac{F_{e}}{k}}*r$    

Now, we need to find $F_{e}$. To do that we have that Fe is equal to the spring force ($F_{k}$):

$F_{e} = F_{k} = -kx$

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:

$F_{k} = -W$

$-k*x = -m*g$

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

$k = \frac{mg}{x} = \frac{1.8 \cdot 10^{-3} kg*9.81 m/s^{2}}{1.2 \cdot 10^{-2} m} = 1.47 N/m$      

Now, we can find the electrostatic force:

$F_{e} = F_{k} = -kx = -1.47 N/m*0.8 \cdot 10^{-2} m = -0.0118 N$

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

$q = \sqrt{\frac{F_{e}}{k}}*r = \sqrt{\frac{0.0118 N}{9 \cdot 10^{9} Nm^{2}/C^{2}}}*4.8 \cdot 10^{-2} m = 54.9 \cdot 10^{-9} C = 54.9 nC$

Therefore, the magnitude of the charge is 54.9 nC.

I hope it helps you!