Question

Two large parallel conducting plates carrying equal and opposite charges on their
surfaces facing each other are placed at a distance of 0.75 m. An electron placed at
some point between the plates experiences an electrostatic force of 4.8 × 10-16 N.
Calculate the magnitude of the electric field at the position of the electron between
the plates. Also determine the potential difference between the plates.

Answer 1

E=F/q

=4.8×10^-16/1.6×10^-19

=3×10^3 N/C

V=d×E

=0.75×3×10^3

=750×3

= 2250 volt

Answer 2

Answer: The electric field is 3000 N/C and the potential difference between the plates is 2250 volt.

Explanation:

Given that,

Distance d = 0.75 m

Electrostatic force $F =4.8\times10^{-16}N$

We know that,

The electric field is

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

Where, $q= 1.6\times10^{-19} C$

$E = \dfrac{4.8\times10^{-16}\ N}{1.6\times10^{-19}\ C}$

$E = 3000\ N/C$

The potential difference is the product of the electric field and the distance between the plates.

$V = E\times d$

$V = 3000\times 0.75$

$V = 2250\ volt$

Hence, The electric field is 3000 N/C and the potential difference between the plates is 2250 volt.