17: Equations Flashcards
Millikan’s oil drop experiment:
Force exerted on a current carrying wire in a magnetic field
F = BIl B = magnetic flux density I = current l = length (m)
equation for current linking velocity of the charge going through the wire
I (current) = qv/l
v = velocity of charged particle
q = charge
l = length
equation for kinetic energy gained by electron between cathode and anode as part of electron deflection
1/2 m v^2 = eV
eV = electronvolts
force due to magnetic field acting on a single charged particle moving through a magnetic field
F = qvB F = force in newtons q = charge in coulombs v = velocity in ms^-1 B = magnetic field strength in teslas
radius of circular path for deflected charged particle
r = mv / qB r = radius in metres m = mass in kg v = velocity in ms^-1 q = charge in coulombs B = magnetic field strength in Teslas
velocity of charged particle with magnetic force and centripetal motion
qvb = mv^2 / r qb = mv / r v = qbr / m
what is the equation to calculate the charge on an oil drop in Millikan’s experiment
q = mgd / v q = charge m = mass in kg d is distance between plates in metres v = voltage
what are the 2 equations that are equal when an oil drop is suspended between 2 charged plates
F = qE, w = mg, qE = mg
what does the equation for electric potential V V = kQ / r, a charged sphere give
V = kQ / r, where r is radial distance from the centre of the charge, which could go on to infinity
gives the work done to bring a unit positive charge from infinity to a given point in an electric field
how to find out electric field strength for electron moving in a straight line through a magnetic field
if its moving in a straight line then force due to electric field must be equal to the force due to magnetic field so Felectric = Eelectric * q F = QvB QVB = EQ so E = VB V = velocity B = magnetic field strength
