Answer to Question #231537 in Electric Circuits for Bus

Question

A proton is placed in a uniform electric field of 3.75 kN/C. Calculate
the magnitude of the electric force felt by the proton, the proton’s
acceleration and the proton’s speed after 1us in the field assuming
initially, it is at rest.

Answer 1

The force is related to the charge of a proton and to the field as $F = qE,$ where E is the field, so $F = 1.6\cdot10^{-19}\,\mathrm{C}\cdot 3.75\cdot10^3\,\mathrm{N/C} = 6\cdot10^{-16}\,\mathrm{N}.$

The acceleration is $a = \dfrac{F}{m} = \dfrac{6\cdot10^{-16}\,\mathrm{N}}{1.67\cdot10^{-27}\,\mathrm{kg}} = 3.6\cdot10^{11}\,\mathrm{m/s^2}.$

After 1 $\mu$s the velocity will be $v = v_0 + at = 0 + at = 3.6\cdot10^{11}\,\mathrm{m/s^2}\cdot 1\cdot10^{-6}\,\mathrm{s} = 3.6\cdot10^5\,\mathrm{m/s}.$