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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.

User Noteness
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1 Answer

11 votes

Answer:

ΔV = - 3274 V

Step-by-step 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 =
(m \ v^2 )/(q)

let's calculate

ΔV =
(4.4 \ 10^(-6) \ 80^2 )/( 2 \ 4.3 10^(-6) )

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

User Anjuman
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