A proton has an initial speed of 3·10 5 m/s. It is brought to rest by an electri

Question

A proton has an initial speed of 3·105 m/s. It is brought to rest by an electric field.

a.) What is the magnitude of the voltage difference needed to bring this charge to rest?
_______ V

b.) At the point when the proton comes to rest (momentarily):
its potential energy is  :
larger / smaller / the same as before / not enough info. ,

and it is at a: larger / smaller / the same as before / not enough info. voltage as compared to initially.

c.) The electric field must be pointed : in the same direction as / in the opposite direction as / perpendicular to / not enough info.  the initial velocity of the proton, so that it is able to stop it.

d.) Describe the motion of the proton after it comes to rest (momentarily).

Explanation / Answer

We use (K+U) = constant where K is kinetic energy = 1/2*m*v^2 and U is potential energy = V*q where V is potential

So (K + V*q)i = (K+V*q)f

a) Here Kf = 0... so 1/2*m*vi^2 = q*(Vf-Vi)...So
(Vf-Vi) = 1/2*m*v^2/q = 1/2*1.67x10^-27kg*(3x10^5m/s)^2/1.602*10^-19 = 469.10 V

b) at rest potential energy will be larger becuase all K.E will be converted into P.E

c) in the opposite direction to stop

d) At rest proton will experience net force = 0, velocity wil be zero.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.