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centers of two small particles charged particles are separated by a distance of 1.2x10-4 meter. The charges on the particle are 8x10-19 coloumb, respectively. Calculate the magnitude of the electrostatic force between these two particles. Please help, I need to get a good grade for physics.

Verified Answer :

dexteright02

Data:
r (distance) = \(1.2*10^{-4}m\)
charges on the particle (Q*q) = \(8*10^{-19}\:Coulomb\)
k (Coulomb\:constant) = \(8.988 * 10^9 N*m2/C2 \to K \approx 9.00 * 10^9 N*m2/C2\)
F (magnitude) = ?

In applying the Coulomb Law we have:
\(F = \frac{k*Q*q}{r^2}\)

Solving:
\(F = \frac{k*Q*q}{r^2}\)
\(F = \frac{9.00 * 10^9*8*10^{-19}}{(1.2*10^{-4})^2}\)
\(F = \frac{72 * 10^{9-19}}{(1.2*10^{-4})^2}\)
\(F = \frac{72 * 10^{-10}}{1.44*10^{-8}}\)
\(F = 50*10^{-10-(-8)}\)
\(F = 50*10^{-10+8}\)
\(F = 50*10^{-2}\)
\(\boxed{\boxed{F = 5.0*10^{-1}\:N}}}\end{array}}\qquad\quad\checkmark\)

Note: Response used with approximate values

asuwafohjames

The magnitude of the electrostatic force between the two particles is 3.99×10⁻²³ N.

To calculate the magnitude of the electrostatic force between the two particles, we use the formula below.

Formula:

  • F = kqQ/r²............... Equation 1

Where:

  • F = Electrostatic force between the two particles
  • q = charge on the first particle
  • Q = charge on the second particle
  • r = distance between the particles.
  • k = Columb's constant

From the question,

Given:

  • Q = q = 8×10⁻¹⁹ C
  • r = 1.2×10⁻⁴ m
  • k = constant = 8.99×10⁹

Substitute the given values into equation 1

  • F = (8×10⁻¹⁹)(8×10⁻¹⁹)(8.99×10⁹)/(1.2×10⁻⁴)
  • F = 399.56×10⁻²⁵
  • F = 3.99×10⁻²³ N

Hence, The magnitude of the electrostatic force between the two particles is 3.99×10⁻²³ N.

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