WEBVTT
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This is College Physics
Answers with Shaun Dychko.
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The size of the change in kinetic
energy of this electron
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is equal to the size of the
change in its potential energy
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which will be its charge times V
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and I am putting these absolute
value bars here to say,
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let’s just ignore negative signs.
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Now, the potential difference for a uniform
electric field is the electric field
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multiplied by the distance
travelled through the field
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and we substitute that in for V
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and this is the same formula we have for
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the potential difference between two
large conducting parallel plates
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and the reason it applies is
because the electric field
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between two large conducting
parallel plates is uniform.
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Okay.
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So… we have a change in energy then is
charge times electric field times distance.
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So, we are dealing with electrons,
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so the charge is the elementary charge of
1.6 times ten to the minus 19 coulombs
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times two times ten to the six volts
per meter electric field strength
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times 0.400 meter distance travelled
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and then we are asked to express our
answers in unit of kilo-electron volt.
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So, this will give us joules
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and we have this conversion factor
to go from joule to electron-volts,
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one electron-volt for every 1.60 times
ten to the minus nineteen joules
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and then we convert that
in to kilo-electron volt
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by multiplying by one kilo-electron
volt for every 1000 electron-volts
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and this works out to
800 kilo-electron volts
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and perhaps I should write the answer
in scientific notation to indicate
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that there are two significant figures…
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oh sorry three significant
figures lets just say, there.
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Okay.
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Now, part B asks us over what distance would
it have to be accelerated in this field
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to gain an energy of 50 Giga-electron volt.
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So, we have q E d equals the
change in kinetic energy,
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that’s what we said up here
and we will solve this for d
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by dividing both sides by q times E.
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So, d is the change in energy over E times q,
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and change in energy is
50 Giga-electron volt
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which we have to convert into joules
in order to use in this formula
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so we multiply by one times ten to the nine
electron-volt for every Giga-electron volt
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and then multiply by 1.60 times ten to the
minus 19 joules for every electron-volt
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and we are left with joules
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and we divide that by electric
field times the only true charge
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and we end up with 25.0 kilometres,
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is the distance over which electron would
need to be accelerate by this field
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to get an energy of 50 giga-electron volt.