Question

The vertical deflecting plates of a typical classroom are a pair of parallel square metal plates carrying equal but opposite charges. Potential difference between the plates is 25.0V; typical dimensions are about 3.8cm on a side with a separation of about 4.6mm. Under these conditions, how much charge is on each plate? How stron is the electric field between the plates? I fan electron is ejected from the negative plates, how fast is it moving when it reaches the postive plate?

Answer 1

`6.945326087* 10^(-11)\ C`

5434.78260873 N/C

2963369.48874 m/s

Step-by-step explanation:

`\epsilon_0` = Permittivity of free space =
`8.85* 10^(-12)\ F/m`

q = Charge of electron =
`1.6* 10^(-19)\ C`

V = Voltage = 25 V

Side = 3.8 cm

d = Separation = 4.6 mm

m = Mass of electron =
`9.11* 10^(-31)\ kg`

Charge is given by

`Q=(VA\epsilon_0)/(d)\\\Rightarrow Q=(25* (3.8* 10^(-2))^2* 8.85* 10^(-12))/(4.6* 10^(-3))\\\Rightarrow Q=6.945326087* 10^(-11)\ C`

Charge on each plate is
`6.945326087* 10^(-11)\ C`

Electric field is given by

`E=(Q)/(A\epsilon_0)\\\Rightarrow E=(6.945326087* 10^(-11))/((3.8* 10^(-2))^2* 8.85* 10^(-12))\\\Rightarrow E=5434.78260873\ N/C`

The electric field strength is 5434.78260873 N/C

Acceleration is given by

`a=(qE)/(m)\\\Rightarrow a=(1.6* 10^(-19)* 5434.78260873)/(9.11* 10^(-31))\\\Rightarrow a=9.5451725291* 10^(14)\ m/s^2`

`v^2-u^2=2as\\\Rightarrow v=√(2as+u^2)\\\Rightarrow v=\sqrt{2* 9.5451725291* 10^(14)* 4.6* 10^(-3)+0^2}\\\Rightarrow v=2963369.48874\ m/s`

The velocity is 2963369.48874 m/s