6) A proton is accelerated through a potential difference of 6 MV. (megavoit fro

Question

6) A proton is accelerated through a potential difference of 6 MV. (megavoit from rest. Calc. the final velocity in m/s (a) 1.4 X 10' (b) 2.4 x 10' (c)3.4X1 (d) 4.4 x 10. 7) Use Gauss' Law to solve this problem. A 6 nC (nanocoulomb) charge is , located at the center of a 2 meter radius uncharged spherical metal shell. Ca the electric field at a radius of r = 1 meter in N/C, (a) 14 (b) 24 (c) 34/(d)54 8) An electric field is 4 X 105 V/m at an angle of 30° to the x axis. A charge undergoes a displacement vector of 3 m j. (3 meters along the y axis). Cal potential difference in volts. (a) 6 x 10 (b) 8 x 10 (c) 1 x 106 (d) 1.2 x 10 12

Explanation / Answer

6)

for proton charge,

q = 1.602*10^-19 C

V = 6 MV = 6*10^6 V

mass of proton,

m = 1.673*10^-27 Kg

so,

work done = q*V

= (1.602*10^-19 C)*(6*10^6 V)

= 9.612*10^-13 J

now use:

work done = charge in kinetic energy

work done = final KE - initial KE

work done = final KE - 0

work done = final KE

work done = 0.5*m*vf^2

9.612*10^-13 = 0.5*(1.673*10^-27)*vf^2

vf^2 = 1.15*10^15

vf = 3.4*10^7 m/s

Answer: c

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