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

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, $F=3.2\times 10^{-15}\ N$

(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 :

$E=\dfrac{F}{q}$

q is the charge on electron

$E=\dfrac{3.2\times 10^{-15}\ N}{1.6\times 10^{-19}\ C}$

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 :

$V=E\times d$

$V=20000\ N\times 0.21\ m$

V = 4200 volts

Hence, this is the required solution.