Question

Two large, parallel, conducting plates are 21 cm apart and have charges of equal magnitude and opposite sign on their facing surfaces. An electrostatic force of 3.2 10-15 N acts on an electron placed anywhere between the two plates.

(Neglect fringing).



(a) Find the magnitude of the electric field at the position of the electron.
______ N/C



(b) What is the potential difference between the plates?
______ V

Answer 1

Step-by-step explanation:

It is given that,

Distance between two conducting plates, d = 21 cm = 0.21 m

Electrostatic force acting on the electron placed anywhere between the two plates,
`F=3.2* 10^(-15)\ N`

(a) Let E is the electric field at the position of the electron. The electric force acting on a charged particle per unit charge is called electric field. It is given by :

`E=(F)/(q)`

q is the charge on electron

`E=(3.2* 10^(-15)\ N)/(1.6* 10^(-19)\ C)`

E = 20000 N

(b) The V is the potential difference between the plates. The relation between the potential difference and the electric field is given by :

`V=E* d`

`V=20000\ N* 0.21\ m`

V = 4200 volts

Hence, this is the required solution.