Question

Two very large parallel metal plates, separated by 0.20 m, are connected across a 12-V source of potential. An electron is released from rest at a location 0.10 m from the negative plate. When the electron arrives at a distance 0.050 m from the positive plate, how much kinetic energy (J) has the electron gained

Answer 1

`{\rm K} = 2.4* 10^(-19)~J`

Step-by-step explanation:

The electric field inside a parallel plate capacitor is

`E = (Q)/(2\epsilon_0 A)`

where A is the area of one of the plates, and Q is the charge on the capacitor.

The electric force on the electron is

`F = qE = (qQ)/(2\epsilon_0 A)`

where q is the charge of the electron.

By definition the capacitance of the capacitor is given by

`C = \epsilon_0(A)/(d) = (Q)/(V)\\(Q)/(\epsilon_0 A) = (V)/(d) = (12)/(0.20) = 60`

Plugging this identity into the force equation above gives

`F = (qQ)/(2\epsilon_0 A) = (q)/(2)((Q)/(\epsilon_0 A)) = (q)/(2)60 = 30q`

The work done by this force is equal to change in kinetic energy.

W = Fx = (30q)(0.05) = 1.5q = K

The charge of the electron is
`1.6 * 10^(-19)`

Therefore, the kinetic energy is
`2.4* 10^(-19)`