Question

proton gun fires a proton from midway between two plates, A and B, which are separated by a distance of 10.9 cm; the proton initially moves at a speed of 113 km/s toward plate B. Plate A is kept at zero potential, and plate B at a potential of 689 V. With what speed will it hit plate A?

Answer 1

potential at any point is proportional to its distance form the 0 potential plate.
V=E*d, where
E - electric field intensity, d-distance from plate 0 plate.
E=V/d=689/0.109=6321.1 V/m

Proton at initial stage has,U= 344.5 V potential.

The energy of proton

m=1.7x10^-27 kg

q=1.6x10^-19 C
v=11.3x10⁴ m/s

E₁=KE+PE=mv²/2 + qU=(1.7x10^-27x(11.3x10⁴)²)/2 +(1.6x10^-19x344.5) = 6.6 x 10^-17 J

When hitting plate A, the proton will have 0 potential energy, therefore everything is stored in speed.

E₂ = mv²/2 + 0

E₂ = E₁

v=278.65 km/s