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the voltage between the anode and the cathode of an electron gun is 2500V. show that the electrons are emitted from the gun at about 3x10^7 m/s, any ideas?
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E = W = QV (volt) = 0.5Mv^2
@goive123 ??
it is using electrong charge and the masss of electrons; i have been given the values
Thus V = 0.5 M v^2 / Q .
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