Respuesta :
Answer:
[tex]{\rm K} = 2.4\times 10^{-19}~J[/tex]
Explanation:
The electric field inside a parallel plate capacitor is
[tex]E = \frac{Q}{2\epsilon_0 A}[/tex]
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
[tex]F = qE = \frac{qQ}{2\epsilon_0 A}[/tex]
where q is the charge of the electron.
By definition the capacitance of the capacitor is given by
[tex]C = \epsilon_0\frac{A}{d} = \frac{Q}{V}\\\frac{Q}{\epsilon_0 A} = \frac{V}{d} = \frac{12}{0.20} = 60[/tex]
Plugging this identity into the force equation above gives
[tex]F = \frac{qQ}{2\epsilon_0 A} = \frac{q}{2}(\frac{Q}{\epsilon_0 A}) = \frac{q}{2}60 = 30q[/tex]
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 [tex]1.6 \times 10^{-19}[/tex]
Therefore, the kinetic energy is [tex]2.4\times 10^{-19}[/tex]
