Answer:
(A) the acceleration experienced by the proton 2.821 x 10¹² m/s²
(B) the speed of the proton is 2.67 x 10⁵ m/s
Explanation:
Given;
electric field experienced by the proton, E = 2.95 x 10⁴ N/C
charge of proton, Q = 1.6 x 10⁻¹⁹ C
mass of proton, m = 1.673 x 10⁻²⁷ kg
distance moved by the proton, d = 1.26 cm = 0.0126 m
(a)
The force experienced by the proton is calculated as;
F = ma = EQ
where;
a is the acceleration experienced by the proton
[tex]a = \frac{EQ}{m} \\\\a = \frac{2.95\times 10^4 \ \times \ 1.6\times 10^{-19}}{1.673 \times 10^{-27}} \\\\a = 2.821 \times 10^{12} \ m/s^2[/tex]
(b) the speed of the proton is calculated;
v² = u² + 2ad
v² = 0 + (2 x 2.821 x 10¹² x 0.0126)
v² = 7.109 x 10¹⁰
v = √7.109 x 10¹⁰
v = 2.67 x 10⁵ m/s
