Three equal 1.60-μC point charges are placed at the corners of an equilateral triangle with sides 0.700 m long. What is the potential energy of the system? (Take as zero the potential energy of the three charges when they are infinitely far apart.)

Respuesta :

Answer:

Potential Energy,[tex]U=0.09874J[/tex]

Explanation:

Given an equilateral triangle with all 3 equal sides and equal charges of say[tex]a_1=a_2=a_3=1.60\mu C[/tex]

Potential energy,U, here is involved in the assembling of charges and the said charges are separated by infinite distances.Total potential energy is the  summation of all potential energy of interaction between each pair of charge.

[tex]U=\frac{1}{4\pi \in_0}(\sum\frac{a_ia_j}{r_i_j})=\frac{1}{4\pi \in_0}(\frac{3a_1^2}{r_1_2})\\\\r_1_2=r_2_3=r_1_3=0.700m\\\\\frac{1}{4\pi \in_0}=9.0*10^9Nm^2/C^2\\\\U=(9.0*10^9Nm^2/C^2)(\frac{3(1.6*10^{-6}*C)^2}{0.700})\\\\U=0.09874J[/tex]

Lanuel

The potential energy of the system is equal to 0.099 Joules.

Given the following data:

  • Quantity of charge = 1.60-μC
  • Radius = 0.700 meter.

Scientific data:

  • Coulomb's constant = [tex]8.99 \times 10^9\;Nm^2/C^2[/tex]

To calculate the potential energy of the system:

How to calculate the total potential energy

The total potential energy (U) of the system is equal to the sum of all potential energy possessed in between each pair of charge.

Mathematically, the total potential energy (U) in an electric field is given by this formula:

[tex]U=\frac{3q^2k}{r^2} [/tex]

Where:

  • k is Coulomb's constant.
  • q is the charge.
  • r is the radius.

Substituting the parameters into the formula, we have;

[tex]U=\frac{3 \times (1.6 \times 10^{-6})^2 \times 8.99 \times 10^9}{0.7^2} \\\\U=\frac{3 \times 2.56 \times 10^{-12} \times 8.99 \times 10^9}{0.49}[/tex]

U = 0.099 Joules.

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