Question

A positively charged plate is parallel to a negatively charged plate. The plates are 4.5 cm from each other, and there is a charge of 120 V across the plates. A positively charged particle is between the plates, 1.0 cm from the negative plate.

a) What is the potential difference at the point of the charged particle?

b) If released from this position, at what speed will the particle hit the negative plate if the particle is a proton?

Answer 1

here,

the separation between the plates , d1 = 4.5 cm

the distance of point charge from negative plate , d2 = 1 cm

potential difference between the plates , V = 120 V

a)

the potential difference at the point of the charged particle and negavtive plated , V2 = Electric feild * d2

V2 = (V/d1) * d2 = 16.67 V

b)

mass of proton , m = 1.67 * 10^-27 kg

charge on a proton , e = 1.6 * 10^-19 C

using work energy theorm

work done by potential = kinetic energy gained

V2 * e = 0.5 * m * v^2

26.67 * 1.6 * 10^-19 = 0.5 * 1.67 * 10^-27 * v^2

solving for v

v = 7.14 * 10^4 m/s

the speed of proton is 7.14 * 10^4 m/s