Question

Question 13 change in potential mengy of a A uniforn electric field, with a magnitude of $00 N C, is directed parallel to the positive z-aris. If the potestial at x -3.0 m is 1,400 V what s the proton as it moves from x-51n to x = 3 m? (@p=16×10-19C) t a 1 12 × 10-21 J 16x 10-17 e 448 × 10-16 J O b 500J 0d488×10-16J 0

Answer 1

C option correct

E = 500 N/C F = q E q = 1.6 times 10^- 19 C - Delta V = -F Delta V Since proton is going opposite to direction of electric field therefore potential energy increases, change in potential energy is positive Delta V = -q E (r_2 - r_1) { r_2 = 3 r_1 = 5 = -500 times 1.6 times 10^- 19 (3 - 5) = -500 times 1.6 times (- 2) times 10^- 19 = 1600 times 10^- 29 J = 16 times 10^- 17 J