2018 10:59 PM 83.3/100 7/5/2018 03:22 PM ?Print ?l calculator -41 Periodic Table

Question

2018 10:59 PM 83.3/100 7/5/2018 03:22 PM ?Print ?l calculator -41 Periodic Table n 28 of 32 Ma Sapling Learning At locations A and B the electric potential has the values 1,19 V and 5,49 V, respectively. You release a proton from rest at one of these locations and it passes through the other one. At which location should you release the particle? What is its speed when it passes through the other location? Number m/s Repeat the same question, but this time using an electron instead of a proton. m/ s O Previous ? Give Up & View Solution O Check Answer Next Exit Hint 6

Explanation / Answer

a)
Since protons are positively charged, it is repelled by positive charges and attracted by negative charges. So it will move from high potential (where electric field lines are more) to low potential
So, proton must start from B.

Electric potential energy is converted into kinetic energy.
qV = 1/2 mpv2
Where q is the charge of proton, V is the potential difference, mp is the mass of the proton and v is the velocity.
v = SQRT[2qV/mp]
= SQRT[2 x (1.602 x 10-19) x (5.49 - 1.19) / (1.673 x 10-27)]
= 2.87 x 104 m/s

b)
As the electrons are negative charged, it will move from a low potential to high potential. So it must start from A.

Electric potential energy is converted into kinetic energy.
qV = 1/2 mev2
Where q is the charge of electron, V is the potential difference, me is the mass of the electron and v is the velocity.
v = SQRT[2qV/me]
= SQRT[2 x (1.602 x 10-19) x (5.49 - 1.19) / (9.11 x 10-31)]
= 1.23 x 106 m/s

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