Question

A uniform electric field of magnitude 624 N/C exists between two parallel plates that are 3.98 cm apart. A proton is released from rest at the positive plate at the same instant an electron is released from rest at the negative plate.

(a) Determine the distance from the positive plate at which the two pass each other. Ignore the electrical attraction between the proton and electron.
m

(b) Repeat part (a) for a sodium ion (Na⁺) and a chloride ion (Cl^?).
cm

Answer 1

(a) By F = qE
=>F = eE = [1.6 x 10^-19] x 624 = 1.03 x 10^-16 N
Thus by a = F/m
=>a(electron) = [1.03 x 10^-16/9.11 x 10^-31] = 1.13 x 10^14 m/s^2
& a(proton) = [1.03 x 10^-16/1.67 x 10^-27] = 6.17 x 10^10 m/s^2
Let the two met after t sec
=>By s = ut + 1/2at^2
=>s(p) = 0 + 1/2 x 6.17 x 10^10 x t^2 -------------(i)
& s(e) = 0 + 1/2 x 1.13 x 10^14 x t^2 -------------(ii)
=>By (i) + (ii) :-
=>4.06 x 10^-6 = 1/2 x 10^10 x t^2 [6.17 + 11300]
=>t = ?[7.18 x 10^-20]
=>t = 2.68 x 10^-10 sec
(a) s(p) = 1/2 x 6.17 x 10^10 x (2.68 x 10^-10)^2 = 2.22 x 10^-9 m
(b) do the same way yourself.