An electron is intially stationary between 2 charged plates at the surface of th

Question

An electron is intially stationary between 2 charged plates at the surface of the negatively charged plate. The plates are 5mm apart and have a potential difference of 20,000V.      A) What is the potential energy of the electron at this position?      B) Wht is the kinetic energy of the electron at its initial position?

If released, the electron will accelerate towards the positivley charged plate.      C) What will lits kinetic energy be when it reaches the positive plate?      D) What will its velocity be?      E) What will its potential energy be when it reaches the plate?

Explanation / Answer

here

part A:

Potential energy U = Vq

U = 20000* 1.6 *10^-19

U = 3.2 *10^-15 Joules

-------------------------

part B :

intial KE = 0.5 mv^2 = 0

as v is zero due to stationary electron

---------------------------------------------

part C:

Gravitational force = electrical force

ma = Eq

a = Eq/m

E = V/d

E = 20000/(5 *10^-3)

E = 4*10^6 V/m

so

accleration a = (4*10^6 * 1.6*10^-19)/(9.11*10^-31)

a = 7*10^17 m/s^2

so final spped V^2 = 2aS

V^2 = 2 * 7*10^17 * 5*10^-3

V = 8.36 *10^7 m/s (answer)

part C: KE = 0.5mv^2

KE = 0.5* 9.11* 10^-31 * 8.36*10^7* 8.36 *10^7

KE = 3.18 *10^-15 Joules

----------------
part D:

Gravitational force = electrical force

ma = Eq

a = Eq/m

E = V/d

E = 20000/(5 *10^-3)

E = 4*10^6 V/m

so

accleration a = (4*10^6 * 1.6*10^-19)/(9.11*10^-31)

a = 7*10^17 m/s^2

so final spped V^2 = 2aS

V^2 = 2 * 7*10^17 * 5*10^-3

V = 8.36 *10^7 m/s (answer)
-----------------------------------------------

PE = change in KE

PE = - 3.18 *10^-15 Joules (answer)

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