Question

An electron is released from rest at the negative plate of a parallel plate capacitor. The charge per unit area on each plate is σ = 1.99 x 10^-7 C/m^2, and the plate separation is 1.69 x 10^-2 m. How fast is the electron moving just before it reaches the positive plate?

Answer 1

Answer:

v = 1.15*10^{7} m/s

Explanation:

given data:

charge/ unit area$= \sigma = 1.99*10^{-7} C/m^2$

plate seperation = 1.69*10^{-2} m

we know that

electric field btwn the plates is$E = \frac{\sigma}{\epsilon}$

force acting on charge is F = q E

Work done by charge q id$\Delta X =\frac{ q\sigma \Delta x}{\epsilon}$

this work done is converted into kinectic enerrgy

$\frac{1}{2}mv^2 =\frac{ q\sigma \Delta x}{\epsilon}$

solving for v

$v = \sqrt{\frac{2q\Delta x}{\epsilon m}$

$\epsilon = 8.85*10^{-12} Nm2/C2$

$v = \sqrt{\frac{2 1.6*10^{-19}1.99*10^{-7}*1.69*10^{-2}}{8.85*10^{-12} *9.1*10^{-31}}$

v = 1.15*10^{7} m/s