Question

two parallel plates separated by the distance of 5 mm are kept in the potential difference of 50 volt a particle of mass 10 ^ - 15 kg and charge of petrol power of minus 11 Columbus internet with the velocity of 10 to the power of minus 3 metre per second the acceleration of the particle will be​

Answer 1

your question ---> two parallel plates separated by a distance of 5 mm are kept at a potential difference of 50V. A particle of mass 10^-15 kg and charge 10^-11 enters in it with a velocity 10^7 m/s. The acceleration of the particle will be?

potential difference between plates ,V= 50 volts

distance between plates, d = 5mm = 5 × 10^-3 m

mass of charged particle ,m = 10^-15 kg

charge on particle , q = 10^-11 C

velocity of particle , v = 10^7 m/s

first of all, we have to find electric field;

electric field, E = V/d =50/5 × 10^-3 = 10000 volts.

now force acting on charged particle q , F = qE = 10^-11 × 10000 = 10^-7 N

from Newton's 2nd law,

F = ma = 10^-7 N

or, 10^-15 × a = 10^-7

or, a = 10^8 m/s²

Acceleration due to gravity , g = 10m/s²

so, acceleration on particle will be √(a² + g²)

= √(10^16 + 10²)

≈ 10^8 m/s²