(a) Determine the acceleration of the proton. (b) Determine the initial speed of

Question

(a) Determine the acceleration of the proton.

(b) Determine the initial speed of the proton.

(c) Determine the time interval over which the proton comes to rest.
? s

magnitude ? m/s2 direction +x ?x +y ?y +z ?z A proton is projected in the positive x direction into a region of uniform electric field vector E= (-6.60 ? 10^5) N/C at t = 0. The proton travels 6.80 cm as it comes to rest. (a) Determine the acceleration of the proton. magnitude ? m/s^2 direction +x ?x +y ?y +z ?z (b) Determine the initial speed of the proton. magnitude ? The electric field is constant, so the force is constant, which means the acceleration will be constant .m/s direction ? +x ?x +y ?y +z ?z (c) Determine the time interval over which the proton comes to rest. ? s

Explanation / Answer

As proton has +ve charge so force on proton will bi in direction of electic fileld ( in tis case -ve x axis)

force on proton = q*E = (-6.60 ? 105)*(+1.6*10-19) = - 1.057*10-13 N in direction of electic field(-ve x axis)

(a)

so acceleration = F/Mp = (1.057*10-13 )/ 1.572*10-27 = 6.3220*1013  m/sec2   (Ans)

(b) V2 =U2 + 2*a*s and V =0

so U = sqrt(2* 6.3220*1013 *0.068) =2932221.00m/sec = 2.932*106 m/sec(approx)

(c) V = U + a*t ans V =0

the t = U/a = 2.932*106 / 6.3220*1013 = 4.638 *10-8 sec = 46.38 ns

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