Question

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 1

Answer:

Explanation:

Electric field, E = 490 N/C

mass of electron, me = 9.1 x 10^-31 kg

mass of proton, mp = 1.67 x 10^-27 kg

charge of electron or proton = 1.6 x 10^-19 C

time, t = 54 ns = 54 x 10^-9 s

initial velocity, u = 0 m/s

Force on each particle, F = q E = 1.6 x 10^-19 x 490 = 7.84 x 10^-17 N

acceleration of electron = Force / mass of electron

ae = (7.84 x 10^-17) / ( 9.1 x 10^-31) = 8.6 x 10^13 m/s²

Let the velocity of electron is ve.

use first equation of motion

ve = u + ae x t

ve = 0 + 8.6 x 10^13 x 54 x 10^-9

ve = 4.65 x 10^6 m/s

acceleration of proton = Force / mass of proton

ap = (7.84 x 10^-17) / ( 1.67 x 10^-27) = 4.69 x 10^10 m/s²

Let the velocity of electron is vp.

use first equation of motion

vp = u + ap x t

vp = 0 + 4.69 x 10^10 x 54 x 10^-9

ve = 2535.1 m/s

Answer 2

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

$a = \dfrac{784\times 10^{-19}}{9.1\times 10^{-31}}$

  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

$a = \dfrac{784\times 10^{-19}}{1.67\times 10^{-27}}$

  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