Okay so EMF and PD can get a little confusing as they’re very similar so i’ll try my best here:
Basically;
Potential difference is defined as the energy transferred from electrical energy to other forms per unit charge. The term is used when charged particles lose energy in a component (when work is being done by the electrons/charge carriers). It is measured in volts. One volt is the potential difference across a component when 1 joule of energy is transferred per unit charge: 1V = 1JC^-1. To determine p.d., use the equation: W=VQ. You can measure p.d. using a voltmeter. They are placed in a parallel formation in a circuit and should have a very very high resistance so no current passes through it when it is placed in the circuit.
Electromotive force is defined as the energy transferred from chemical energy (or another form) into electrical energy per unit charge. The term is used when charged particles gain energy (when work is being done on the electrons/charge carriers). It also is measured in volts. You also use a voltmeter to measure it. You also use W=VQ to measure it (but in this instance, it is W=EQ, where E is emf). See what I mean….they’re pretty similar:/
Nonetheless we can learn it ahaha. The energy transferred to or from the charges can be calculated using W=VQ=EQ. The amount of energy transferred depends on the size of the charge passing through the component and also the size of the p.d or emf.
Okay so there’s this bit about eV=0.5mv^2 in the spec that best fits in here if I explain the electron gun, so that’s what i’ll do…
Basically in an electron gun (a device that produces a narrow beam of electrons) a metal filament is heated by an electric current and the electrons gain kinetic energy (well, first they gain thermal energy then it turns into kinetic energy). Some gain enough kinetic energy to escape from the metal (thermionic emission). The anode has a small hole in it. The escaped electrons accelerate towards the anode (gaining more kinetic energy as they do). The electrons in line with the hole pass through, creating a beam of electrons with a specific kinetic energy.
The work done on an electron travelling in this beam (p.d.) is equal to e x V (eV, but do not get this mixed up with electron volts! that is a different thing, in this instance eV means elementary charge x the accelerating p.d.). The work done on the electron equates to its gain in kinetic energy. Therefore...
...provided the electrons have negligible kinetic energy at the cathode.
This means that the greater the p.d (eV), the greater the kinetic energy of the electrons.
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