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Higher Unit 2 - The Energy Confusion

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    Posted by BitesizePhysicsTeacher (U15165682) on Tuesday, 15th January 2013

    Unit 2 in Higher has two situations where the energy of something electrical is of interest.

    These two situations are often confused as they both involve energy, charge and voltage (or potential difference).

    It is vital that you recognise the differences between the two situations and don't just go blindly throwing numbers into equations hoping that you're doing the right thing.

    Situation 1 - Work done on a charge in an electrical field

    W = qV (or E = qV) - here, W (E) is the work done (i.e. energy transferred) in accelerating a charged particle - most commonly an electron in a cathode ray tube.

    This energy is transferred to kinetic energy, so that -

    work done, W = E(k), kinetic energy gained
    qV = ½ m v²

    Often you'll be asked to find the 'work done on', 'energy transferred to' or 'energy gained by' the charged particle - these are all the same and found using W = qV

    You may be asked to find the speed of the accelerated particle, in which case,

    v = √(2E/m) or v = √(2qV/m)

    More detail on this can be found on Bitesize here -

    Situation 2 - Energy stored in a capacitor

    E = ½ Q V = ½ C V² = ½ Q² / C - here E is the electrical potential energy stored in a capacitor.

    The capacitance, C, of a capacitor is a measure of the quantity of charge, Q, that it can store at a given potential difference (or voltage), V, so that C = Q / V.

    As charge, Q, builds up in a capacitor the potential difference, V across it increases. But V = E / Q, the energy per unit charge, which must also be increasing as the charge builds up. This makes sense as the charges building up will repel each other, making it more difficult to add more charge still, so more work will need to be done to add more charge.

    If the voltage isn't constant, then it's best to think of the energy stored as the area under a graph of charge versus potential difference - - as this is a triangle its area is ½ x base x height, so E = ½ Q V.

    For a capacitor, C = Q / V, which allows us to substitute for Q or V in the equation above, giving E = ½ C V² and E = ½ Q² / C

    The energy stored in a capacitor remains there until the capacitor is discharged, when it will be transferred in other forms of energy in an external circuit - e.g. transferred to light in the flash of a camera.

    There's more detail about this on Bitesize, here -

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