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A particle with a charge of -4.3 μC and a mass of 4.4 x 10-6 kg is released from rest at point A and accelerates toward point B, arriving there with a speed of 80 m/s. The only force acting on the particle is the electric force. What is the potential difference VB - VA between A and B? If VB is greater than VA, then give the answer as a positive number. If VB is less than VA, then give the answer as a negative number.

Answer :

Answer:

ΔV = - 3274 V

Explanation:

For this exercise we can use conservation of energy

starting point.

         Em₀ = U = q ΔV

final point

          Em_f = K = ½ m v²

energy is conserved

          Em₀ = Em_f

          q ΔV = ½ m v²

          ΔV = [tex]\frac{m \ v^2 }{q}[/tex]

let's calculate

          ΔV = [tex]\frac{4.4 \ 10^{-6} \ 80^2 }{ 2 \ 4.3 10^{-6} }[/tex]

          ΔV = 3274.4 1 V

since the charge q is negative, the potential at point B must be less than the potential at point A, so the answers

          ΔV = - 3274 V

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