What is the Net Force of the applied forces? *
1 answer:
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GIven data:
Distance between the plates = 1.5 mm
Potential difference V = 600V
Charge on electron q = -1.6× C
mass on electron = m = 9.1× Kg
Solution:
First we will find the change in potential energy of the charge while moving through the potential difference of 600V.
ΔU = qΔV
= (-1.6×)(600)
= -9.6×J
By the law of conservation of mechanical energy, as there is no external force acting, so the sum of the kinetic and potential energies will be a constant.
K + U = E
ΔK + ΔU = 0
ΔK = -ΔU
1/2mv² = -ΔU
v² = -2ΔU/m
=
v =
v = 1.45× m/s