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 = 2.2 × 10^-7 C/m^2, and the plates are separated by a distance of 1.3 × 10^-2 m. How fast is the electron moving just before it reaches the positive plate?

Answer 1

Answer:$1.066\times 10^7 m/s$

Explanation:

Given

Charge per unit area on each plate($\sigma$)=$2.2\times 10^{-7}$

Plate separation(y)=0.013 m

and velocity is given by

$v^2-u^2=2ay$

where a=acceleration is given by

$a=\frac{F}{m}=\frac{eE}{m}$

e=charge on electron

E=electric field

m=mass of electron

$E=\frac{\sigma }{\epsilon _0}$

$a=\frac{e\sigma }{m\epsilon _0}$

substituting values

$v=sqrt{\frac{2e\sigma y}{m\epsilon _0}}$

$v=\sqrt{\frac{2\times 1.6\times 10^{-19}\times 2.2\times 10^{-7}\times 0.013}{9.1\times 10^{-31}\times 8.85\times 10^{-12}}}$

$v=1.066\times 10^7 m/s$