Question

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 1

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