Question

A thin conducting square plate 1.0 m on the side is given a charge o m/s2) magnitude direction 2.0 × 10 。C. A proton is placed 1.0 cm above the center o the plate what is the acceleration o the proton? Enter the magnitude in toward the plate

Answer 1

Here ,for the electric field ,

electric field = charge density/(epsilon)

Electric field = 2 *10^-6/(8.854 *10^-12)

Electric field = 225886.6 N/C

let the acceleration is a

m * a =q * E

1.67 *10^-27 * a = 225886.6 * 1.602 *10^-19

solving for a

a = 2.17 *10^13 m/s^2

the acceleration of the proton is 2.17 *10^13 m/s^2 towards the plate