Electrical Potential and Conservation of Energy?

Question

A proton is released from rest in a uniform electric field of magnitude 70000 V/m directed along the positive x axis. The proton undergoes a displacement from A to B of 0.2 m in the direction of the electric field. Ke = 8.98755e9 Nm^2/C^2 and the mass of a proton is 1.672623e-27 kg.

A) Find the change in the electric potential if the proton moves from the point A to B (answer in V).

B) Find the change in potential energy of the proton for this displacement (answer in J).

C) Apply the principal of energy conservation to find th speed of the proton after it has moved 0.2 m, starting from rest (answer in m/s).

Answer 1

A) Change in electric potential DV = 70000 (V/m) * 0.2 (m) = 14000 (V)
B) Change in potential energy (proton charge Q = 1.602E-19 (As))
DE = Q*DV = 1.602E-19 (As) * 14000 (V) = 2.2428E-15 (J)
C) Due to energy conservation law: mv^2/2 = DE. Since proton mass m = 1.6726E-27 (Kg) follows v = 1637624.116 m/sec.

Answer 2

the flair potential on the alpha particle is fantastic*2*e²/(?[x² + y²] + ok*2*e²/?[x² + y²] = ok*4*e²/(x*?2), the place x = y = 2.ninety two*10^-15 m. the flair potential on the protons is fantastic*e²/(5.fifty 4*10^-15) + ok*2*e²/(x*?2). the flair potential of the alpha particle at ? is 0, so the replace = the flair potential in its preliminary place (calculated above). to locate the fee, equate that to 0.5*m*v² and remedy for v, the place m = mass of the alpha particle the flair potential of the protons are equivalent and replaced into calculated above; back equate this to 0.5*m*v² and remedy for v, utilising m because of the fact the mass of the proton.