Problem 7 a) A proton with kinetic energy of 1.25 x 10 eV is fired perpendicular

Question

Problem 7 a) A proton with kinetic energy of 1.25 x 10 eV is fired perpendicular to the face of a large plate that has a uniform charge density of +6.75 x 10-06 C/m2. What is the magnitude of the force on the proton? Submit Answer Tries 0/6 b) How much work must the electric field do on the proton to bring it to rest? Give only the magnitude of the work done. Submit Answer Tries 0/6 c) From what distance should the proton be fired so that it stops right at the surface of the plate? Submit Answer Tries 0/6

Explanation / Answer

(a) The expression for the strength of homogeneous electric field of on the plate is given by -

E=0.5*p/eps, where given p = 6.75 x 10^-6 C/m^2 and eps = 8.854 x 10^-12

therefore, a proton with kinetic energy w = 1.25 x 10^6 eV = 1.25 x 10^6 * 1.6 x 10^-19 J will undergo a force F=e*E decelerating it, where e=1.602 x 10^-19 C.

So, F = e*E = 1.602 x 10^-19 x 0.5 x 6.75x 10^-6 / (8.854 x 10^-12) = 0.6106 x 10^-13 = 6.106 x 10^-14 N

(b) work = W = 1.25 x 10^6 x 1.6 x 10^-19 = 2.0 x 10^-13 Joules

(c) work done by the field is w= F*x, where x is the distance in question;

therefore, x= W / F =(2.0 x 10^-13) / (6.106 x 10^-14) = 3.27 m

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