Question

A proton moves through an electric potential created by a number of source charges. Its speed is 2.5 x 105 m/s at a point where the potential is 1500 V. What will be the proton’s speed a short time later when it reaches a point where the potential is −500 V? The mass of a proton is 27 m 1.67 10 kg.

Answer 1

Answer:

Proton’s speed, a short time later when it reaches a point of lower potential is 1.4 x 10⁵ m/s

Explanation:

Given;

initial speed of proton, u = 2.5 x 10⁵ m/s

initial potential, V = 1500 V

mass of proton = 1.67 x 10⁻²⁷ kg

Work done, W = eV= ΔK.E = ¹/₂mu²

eV = ¹/₂mu² (J)

where;

e is the charge of the proton in coulombs

V is the electric potential in volts

m is the mass of the proton in kg

u is the speed of the proton in m/s

$m =\frac{2eV_1}{u_1{^2}} = \frac{2eV_2}{u_2{^2}} = \frac{V_1}{u_1{^2}}} =\frac{V_2}{u_2{^2}}$

$\frac{V_1}{u_1{^2}}} =\frac{V_2}{u_2{^2}} = \frac{1500}{(2.5*10^5)^2} = \frac{500}{u_2{^2}} \\\\u_2{^2} =\frac{500*(2.5*10^5)^2}{1500} = 0.333*6.25*10^{10}\\\\u_2 = \sqrt{0.333*6.25*10^{10}} =1.4 *10^5 \ m/s$

Therefore, proton’s speed a short time later when it reaches a point of lower potential is 1.4 x 10⁵ m/s