Question

Example 2 A proton is released from rest at point A in a uniform electric field that has a magnitude of 8.0 × 104 V/m. The proton undergoes a displacement of magnitude d- 0.50 m to point B in the direction of the electric field. Find the speed of the proton after completing the displacement

Answer 1

Force on proton = q_p E q_p (charge on proton) = 1.6 times 10^-19 C F = 1.6 times 10^-19 times 8 times 10^4 F = 1.28 times 10^-14 N acceleration = F/mp = 1.67 times 10^-27 kg a = 1.28 times 10^-14/1.67 times 10^-27 a = 7.65 times 10^12 m/s^2 s = 0.50 m By motion equation = v^2 = u^2 + 2as v^2 = 2 times 7.65 times 10^12 times 0.5 v = 2.766 times 10^6 m/s