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What is the force between two 1.0 x 10-5 c charges separated by 2.0 m?

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Nuhulawal20idn

The force between the two identical electrical charges with the given distance of seperation is 0.2247N.

Given the data in the question;

  • Two Charge; [tex]q_1\ and \ q_2 = 1.0* 10^{-5}C[/tex]
  • Distance between [tex]q_1\ and \ q_2[/tex]; [tex]r = 2.0m[/tex]
  • Force between [tex]q_1\ and\ q_2[/tex]; [tex]F = ?[/tex]

Coulomb's law

Coulomb's law states that the force between two electrical charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between the two charges.

This is expressed as;

[tex]F = k\frac{q_1q_2}{r^2}[/tex]

Where [tex]q_1\ and \ q_2[/tex] are the two electrical charges, r is the distance between the charges and k is the Coulomb constant ( [tex]k = 8.98755*10^9 kgm^3/s^2C^2[/tex] )

To determine the force between the two charges, we substitute our values into the expression above.

[tex]F = k\frac{q_1q_2}{r^2}\\\\F = (8.98755*10^9 kgm^3/s^2C^2)\frac{(1.0*10^{-5}C)(1.0*10^{-5}C)}{(2.0m)^2}\\\\F = (8.98755*10^9 kgm^3/s^2C^2)\frac{1.0*10^{-10}C^2}{4.0m^2}\\\\F = \frac{0.898755kgm^3/s^2}{4.0m^2}\\ \\F = 0.2247kgm/s^2\\\\F = 0.2247N[/tex]

Therefore, the force between the two identical electrical charges with the given distance of seperation is 0.2247N.

Learn more about coulomb's law: https://brainly.com/question/506926

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