Question

two parallel plates separated by a distance of 1cm have a potential difference of 20Vb/w them the plates are held at the horizontal position with negative plate above the positive plate an electron is released from rest in the upper plate what is the acceleration of the electron

Answer 1

Given that,

Potential difference = 20 V

Distance = 1 cm

We need to calculate the acceleration of the electron

Using formula of electric field

$F = qE$

$ma=\dfrac{qV}{d}$

$a=\dfrac{qV}{md}$

Where, m = mass of electron

q = charge of electron

V = potential

d = distance

Put the value into the formula

$a=\dfrac{1.6\times10^{-19}\times20}{9.1\times10^{-31}\times1\times10^{-2}}$

$a=3.51\times10^{14}\ m/s^2$

Hence, The acceleration of the electron is $3.51\times10^{14}\ m/s^2$