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What is the potential difference VB − VA of a particle of mass and charge. Physics Electricity and Magnetism?
A particle of mass 9 × 10−5 kg and charge 2 µC moves downward from point A to point B a distance of 4 m in the earth’s gravitational field. The kinetic energy of the particle decreases by 0.6 mJ during this movement.
What is the potential difference VB − VA? The acceleration of gravity is 9.8 m/s^2.
1 Answer
- Steve4PhysicsLv 73 years agoFavorite Answer
KE = kinetic energy. GPE = gravitational potential energy. EPE = electric potential energy.
ΔV = VB – VA
_______________________
Conservation of energy tells us:
(change in kinetic energy) + (change in GPE) + (change in EPE) = 0
Which is simply saying the loss in kinetic energy equals the overall gain in potential energy.
Δ(KE) + Δ(GPE) + Δ(EPE) = 0 (equation 1)
Take care with signs:
Δ(KE) = -0.6x10^-3J (negative because we are told KE decreases)
Δ(GPE) = mgΔh = 9×10^−5 x 9.8 x (-4) = -0.003528J (negative because height has decreased)
Δ(EPE) = qΔV = 2x10^-6ΔV
From equation 1:
(-0.6x10^-3) + (-0.003528) + (2x10^-6ΔV) = 0
2x10^-6ΔV= 0.004128
ΔV = 2064 volts
. . = 2.1x10^3V to 2 sig. figs.