A proton with an initial speed of 800,000 m/s is brought to rest by an electric

Question

A proton with an initial speed of 800,000 m/s is brought to rest by an electric field. Did the proton move into a region of higher potential or lower potential? A) Because the proton is a positive charge and it slows down as it travels, it must be moving from a region of higher potential to a region of lower potential. B) Because the proton is a negative charge and it accelerates as it travels, it must be moving from a region of lower potential to a region of higher potential. C) Because the proton is a positive charge and it slows down as it travels, it must be moving from a region of lower potential to a region of higher potential. D) Because the proton is a negative charge and it accelerates as it travels, it must be moving from a region of higher potential to a region of lower potential.

What was the potential difference that stopped the proton?

What was the potential difference that stopped the proton?

What was the initial kinetic energy of the proton, in electron volts?

Explanation / Answer

1. Since the kinetic energy decreased, the potential energy must have increased. The proton has a positive charge, so this is an increase in electric potential, i.e. moving to a region of higher potential.

U = q*dV

dV = +ve means lower potential to higher potential

Ans is option C.

2. using enrgy conservation:

KE1 + PE1 = KE2 + PE2

0.5*mv^2 = 0 + PE2 - PE1

dU = KEi = 0.5*1.67*10^-27*(8*10^5)^2

dU = q*dV = 5.344*10^-16

dV = 5.344*10^-16/(1.6*10^-19)

dV = 3340 V

3. KEi = 5.344*10^-16 J

5.344*10^-16/(1.6*10^-19) = 3340 eV

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