Answer:
the acceleration of the proton is 6.025 x 10¹⁰ m/s².
Explanation:
Given;
magnitude of electric field, E = 630 N/C
final speed of the proton, v = 1.5 M m/s = 1.5 x 10⁶ m/s
charge of proton, Q = 1.6 x 10⁻¹⁹ C
mass of proton, m = 1.673 x 10⁻²⁷ kg
The force experienced by the proton is calculated as;
[tex]F = ma = EQ\\\\a = \frac{EQ}{m} \\\\a = \frac{(630)(1.6\times 10^{-19})}{1.673 \times 10^{-27}} \\\\a = 6.025 \times 10^{10} \ m/s^2[/tex]
Therefore, the acceleration of the proton is 6.025 x 10¹⁰ m/s².
