In a region of space, a uniform electric field of 475 N/C points in the positive

Question

In a region of space, a uniform electric field of 475 N/C points in the positive x direction. a) For a proton released from rest in that field, calculate i. the magnitude and direction of the force exerted on it by the field ii. The magnitude and direction of the resulting acceleration iii. its speed 2.50 ns after it is released b) For an electron released from rest in that field, calculate i. the magnitude and direction of the force exerted on it by the field ii. The magnitude and direction of the resulting acceleration iii. its speed 2.50 ns after it is released

Explanation / Answer

a) (i) Fe = q E
  
Fe = (1.6 x 10^-19) (475) = 7.6 x 10^-17 N in the +ve x direction.

(ii) a =Fe / m = 7.6 x 10^-17 / (1.67 x 10^-27 )

= 4.55 x 10^10 m/s^2 in the +ve x direction

(iii) v = u + at

v = 0 + (4.55 x 10^10)(2.50 x 10^-9) = 113.77 m/s

(B) (i) Fe = q E
  
Fe = (- 1.6 x 10^-19) (475) = - 7.6 x 10^-17 N in the +ve x direction.

So 7.6 x 10^-17 N in the -ve x direction.

(ii) a =Fe / m = 7.6 x 10^-17 / (9.109 x 10^-31 )

= 8.34 x 10^13 m/s^2 in the -ve x direction

(iii) v = u + at

v = 0 + (8.34 x 10^13)(2.50 x 10^-9) = 208584.92 m/s

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