Question

An electron is projected horizontally into the space between two oppositely charged metal plates. The electric field between the plates is 557.0 N/C, directed upward.

(a) While in the field, what is the force on the electron?
(b) If the vertical deflection of the electron as it leaves the plates is 3.00 mm, how much has its kinetic energy increased due to the electric field?

Answer 1

F = -8.91 x 10⁻¹⁷ N

`\Delta KE = 2.67* 10^(-19)\ J`

Step-by-step explanation:

given,

Electric field, E = 557.0 N/C

charge of electron,q = -1.6 x 10⁻¹⁹ C

a) Force on the electron

we know,

F = q E

F = -1.6 x 10⁻¹⁹ x 557.0

F = -8.91 x 10⁻¹⁷ N

negative sign of force indicate that force is acting in downward direction.

b) deflection of electron,d =3 mm = 0.003 m

`\Delta KE = W`

`\Delta KE = F.d`

`\Delta KE = -8.91* 10^(-17)* 0.003`

`\Delta KE = 2.67* 10^(-19)\ J`

Increase in Kinetic energy is equal to
`\Delta KE = 2.67* 10^(-19)\ J`