Question

A proton and an electron are moving due east in a constant electric field that also points due east. The electric field has a magnitude of 7.0 x 104 N/C. Determine the magnitude of the acceleration of the proton and the electron. m/sn2 m/s 2 4

Answer 1

Here

for the acceleration of proton is ap

Using second law of motion

1.67 *10^-27 * a = 1.602 *10^-19 * 7 *10^4

solving for a

a = 6.71*10^12 m/s^2

the magnitude of acceleration of proton is 6.71*10¹² m/s^2

for the electron

Using second law of motion

9.11 *10^-31 * a = 1.602 *10^-19 * 7 *10^4

solving for a

a = 1.23*10^16 m/s^2

the magnitude of acceleration of electron is 1.23*10¹⁶ m/s^2