Question

A proton accelerates from rest in a uniform electric field of 630 N/C. At one later moment, its speed is 1.50 Mm/s (nonrelativistic because v is much less than the speed of light). (a) Find the acceleration of the proton.

Answer 1

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;

$F = ma = EQ\\\\a = \frac{EQ}{m} \\\\a = \frac{(630)(1.6\times 10^{-19})}{1.673 \times 10^{-27}} \\\\a = 6.025 \times 10^{10} \ m/s^2$

Therefore, the acceleration of the proton is 6.025 x 10¹⁰ m/s².