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A small particle (m= 2.00 x 10⁴ kg) is charged with -5.00 uC. As it passes point A, where the potential is VA = +200 V, it is moving at 5.00 m/s. What is its speed as it passes point B, which is at a potential V8 = +800 V?

Answer :

Final answer:

The speed of the particle as it passes point B, which is at a potential of +800 V, is approximately 4.97 m/s.

Explanation:

To find the speed of the particle as it passes point B, we can use the principle of conservation of energy. The work done by the electric field on the particle is equal to the change in kinetic energy of the particle.

Given:

  • Mass of the particle (m) = 2.00 x 10⁴ kg
  • Charge of the particle (q) = -5.00 uC = -5.00 x 10⁻⁶ C
  • Potential at point A (VA) = +200 V
  • Initial speed of the particle at point A (vA) = 5.00 m/s
  • Potential at point B (VB) = +800 V

First, let's calculate the work done by the electric field on the particle as it moves from point A to point B:

ΔV = VB - VA = +800 V - (+200 V) = +600 V

W = qΔV = (-5.00 x 10⁻⁶ C)(+600 V) = -3.00 x 10⁻³ J

Since the work done by the electric field is equal to the change in kinetic energy, we can equate the work done to the change in kinetic energy:

W = ΔKE

-3.00 x 10⁻³ J = ½mvB² - ½mvA²

Substituting the given values:

-3.00 x 10⁻³ J = ½(2.00 x 10⁴ kg)vB² - ½(2.00 x 10⁴ kg)(5.00 m/s)²

Simplifying the equation:

-3.00 x 10⁻³ J = ½(2.00 x 10⁴ kg)vB² - ½(2.00 x 10⁴ kg)(25.00 m²/s²)

-3.00 x 10⁻³ J = 10⁴ kgvB² - 2.50 x 10⁵ kg·m²/s²

10⁴ kgvB² = -3.00 x 10⁻³ J + 2.50 x 10⁵ kg·m²/s²

10⁴ kgvB² = 2.47 x 10⁵ kg·m²/s²

vB² = (2.47 x 10⁵ kg·m²/s²) / (10⁴ kg)

vB² = 24.7 m²/s²

vB = √(24.7 m²/s²) = 4.97 m/s

Therefore, the speed of the particle as it passes point B is approximately 4.97 m/s.

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