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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?

Respuesta :

whitneytr12

Answer:

The magnitude of the charge is 54.9 nC.

Explanation:

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

[tex] F_{e} = \frac{k*q_{1}q_{2}}{r^{2}} [/tex]

Where:

q₁ and q₂ are the charges, q₁ = q₂  

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

[tex]F_{e}[/tex]: is the electrostatic force

[tex] F_{e} = \frac{k*q^{2}}{r^{2}} \rightarrow q = \sqrt{\frac{F_{e}}{k}}*r [/tex]    

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

[tex] F_{e} = F_{k} = -kx [/tex]

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:

[tex] F_{k} = -W [/tex]

[tex] -k*x = -m*g [/tex]

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

[tex] 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 [/tex]      

Now, we can find the electrostatic force:

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

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

[tex]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[/tex]

Therefore, the magnitude of the charge is 54.9 nC.

I hope it helps you!  

Lanuel

The magnitude of the charge (in nC) on each bead is equal to 55.21 nC.

Given the following data:

  • Original length = 4.0 cm to m = 0.04 m
  • Mass = 1.8 grams to kg = 0.0018
  • New length = 5.2 cm to m = 0.052.
  • Final length = 4.8 cm to m = 0.048 m.

Extension, e = [tex]0.052 - 0.048[/tex] = 0.012 m

Scientific data:

  • Acceleration due to gravity = 9.8 [tex]m/s^2[/tex]
  • Coulomb's constant = [tex]8.99 \times 10^9\; Nm^2/C^2[/tex]

To calculate the magnitude of the charge (in nC) on each bead, we would apply Coulomb's law:

First of all, we would determine the spring constant of this lightweight spring by using this formula:

[tex]W = mg = Ke \\\\K=\frac{mg}{e} \\\\K=\frac{0.0018 \times 9.8}{0.012} \\\\K=\frac{0.01764}{0.012}[/tex]

Spring constant, K = 1.47 N/m.

For the electrostatic force:

[tex]F = ke\\\\F = 1.47 \times 0.08[/tex]

F = 0.01176 Newton.

Coulomb's law of electrostatic force.

Mathematically, the charge in an electric field is given by this formula:

[tex]q = \sqrt{\frac{F}{k} } \times r[/tex]

Substituting the given parameters into the formula, we have;

[tex]q = \sqrt{\frac{0.01176 }{8.99 \times 10^9} } \times 0.048\\\\q=\sqrt{1.3228 \times 10^{-12}} \times 0.048\\\\q=1.1502 \times 10^{-6} \times 0.048\\\\q= 5.521 \times 10^{-8}\;C[/tex]

Note: 1 nC = [tex]1 \times 10^{-9}\;C[/tex]

Charge, q = 55.21 nC.

Read more on electric field here: https://brainly.com/question/14372859

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