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

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

Step-by-step explanation:

given data:

charge/ unit area
`<strong> = \sigma = 1.99*10^(-7) C/m^2</strong>`

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

we know that

electric field btwn the plates is
`E = (\sigma)/(\epsilon)`

force acting on charge is F = q E

Work done by charge q id
`\Delta X =( q\sigma \Delta x)/(\epsilon)`

this work done is converted into kinectic enerrgy

`(1)/(2)mv^2 =( q\sigma \Delta x)/(\epsilon)`

solving for v

`v = \sqrt{(2q\Delta x)/(\epsilon m)`

`\epsilon = 8.85*10^(-12) Nm2/C2`

`v = \sqrt{(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