A proton is placed at rest near the positive side of two oppositely charged, parallel plates. If there is a potential difference of 12 V, what is the final speed of the proton just before it strikes the other plate? Ans. 47,952 m/s
How do I get that answer?
How do I get that answer?
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Think about it, it's asking for speed of a moving object, so it has energy; that is to say kinetic energy.
Now K.E. = 1/2(mv^2)
Nevertheless its energy and E = qV
So think about it in these terms:
E = 1/2(mv^2)
So v=sqrt((2E)/m)
Substitute for E:
v=sqrt((2qV)/m)
We know V = 12V, q(charge of proton) = 1.6x10^-19 C
m(proton)= 1.67x10^-27 kg so
v=sqrt((2*1.6x10^-19 C*12V)/1.67x10^-27 kg)
apx:
v= 47,952
Now K.E. = 1/2(mv^2)
Nevertheless its energy and E = qV
So think about it in these terms:
E = 1/2(mv^2)
So v=sqrt((2E)/m)
Substitute for E:
v=sqrt((2qV)/m)
We know V = 12V, q(charge of proton) = 1.6x10^-19 C
m(proton)= 1.67x10^-27 kg so
v=sqrt((2*1.6x10^-19 C*12V)/1.67x10^-27 kg)
apx:
v= 47,952
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Never mind. You have to sqrt it last. Thank you !
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