An electron and a proton are each placed at rest in an electric field of 490 n/c. calculate the speed (and indicate the direction) of each particle 54.0 ns after being released.

Answer

given,

Electric field,E = 490 N/C

time, t = 54 ns

for electron

Mass of electron me = 9.1 x 10⁻³¹ kg

Charge of electron e = -1.6 x 10⁻¹⁹  C

electrostatic force

F = E q

F = 490 x 1.6 x 10⁻¹⁹

F = 784  x 10⁻¹⁹ N

now, using newton second law

  a = 8.62 x 10¹² m/s²

using equation of motion

v = u + a t

v = 0 + 8.62 x 10¹² x 54 x 10⁻⁹

v = 4.65 x 10⁵ m/s

velocity of electron is equal to v = 4.65 x 10⁵ m/s

For Proton  

Mass mp = 1.67 x 10⁻²⁷ kg  

Charge p = 1.6 x 10⁻¹⁹ C

Electric field E = 490 V/C

from above solution

F = 784  x 10⁻¹⁹ N

now, acceleration

  a = 4.69 x 10¹⁰ m/s²

using equation of motion

v = u + a t

v = 0 + 4.69 x 10¹⁰ x 54 x 10⁻⁹

v = 4.65 x 10³ m/s

velocity of electron is equal to v = 4.65 x 10³ m/s