A proton accelerates from rest in a uniform electric field of 610 N/C. At some l

Question

A proton accelerates from rest in a uniform electric field of 610 N/C. At some later time, its speed is 1.02-106 m/s. (a) Find the magnitude of the acceleration of the proton. 61-109 Calculate the force that the electric field exerts on the proton and use that force to calculate the acceleration of the proton.m/s2 (b) How long does it take the proton to reach this speed? 173 You appear to have made an error in converting units or else in keeping track of exponents. s (c) How far has it moved in that interval? (d) What is its kinetic energy at the later time? Need Help? Read ItMaster

Explanation / Answer

1.

Using force balance on proton:

Fe = Fnet

q*E = m*a

a = q*E/m

m = mass of proton

Using given values

a = (1.6*10^-19 C)*(610 N/C)/(1.67*10^-27 kg)

a = 5.84*10^10 m/s^2

2.

Given that

U = 0 = initial speed

V = 1.02*10^6 m/sec

a = 5.84*10^10 m/s^2

So Using equation

V = U + a*t

t = (1.02*10^6 - 0)/(5.84*10^10)

t = 1.75*10^-5 sec = 17.5 us

Part C.

distance travled will be given by:

S = U*t + 0.5*a*t^2

S = 0*1.75*10^-5 + 0.5*5.84*10^10*(1.75*10^-5)^2

S = 8.94 m

4.

Kinetic energy will be

KE = 0.5*m*V^2

KE = 0.5*1.67*10^-27*(1.02*10^6)^2

KE = 8.69*10^-16 J

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