Two parallel plates having charges of equal magnitude but opposite sign are sepa

Question

Two parallel plates having charges of equal magnitude but opposite sign are separated by 17.0 cm. Each plate has a surface charge density of 30.0 nC/m2. A proton is released from rest at the positive plate.

(a) Determine the potential difference between the plates.
_______V

(b) Determine the kinetic energy of the proton when it reaches the negative plate.
__________J

(c) Determine the speed of the proton just before it strikes the negative plate.
____________km/s

(d) Determine the acceleration of the proton.
___________m/s2 (towards the negative plate)

(e) Determine the force on the proton
____________N (towards the negative plate.)

(f) From the force, find the magnitude of the electric field.
___________kN/C

Explanation / Answer

here,

distance between the plates, d = 17 cm = 0.17 m
charge density of plates, sigma = 30 nC/m^2 = 30*10^-9 C/m^2

Part A:
potential difference on plates, v

Vp = E*d
Vp = (sigma/eo)*d
Vp = ( (30*10^-9)/(8.85*10^-12) ) / 0.17
Vp = 1.994* 10^4 V

Part 2:
From Conservation of Energy,
Ek = q*Vp
Ek = 1.9*10^-19 * 1.994* 10^4
Ek = 3.789*10^-15 J

Part 3:
Kinetic Energy of proton,
KE = 0.5 * mass of proton * (Velocity)^2

Solving for velocity of proton before collision, v

v = sqrt(2*KE/m)
v = sqrt(2*Ek/m) ( From cons. of energy, Ek = KE)
v = sqrt( (2*3.789*10^-15) / (1.6726*10^-27) )
v = 2.129*10^6 m/s or 2129 km/s

Part 4:
fron newton law of motion,
Force = mass * Acceleration
F = ma

also,
Electric force, F = q*E = q*sigma/eo

therefore

ma = q*sigma/eo

solving for acceleration of proton, a

a = q*Sigma/m*eo
a = (1.6*10^-19*30*10^-9)/(1.6726*10^-21*8.85*10^-12)
a = 3.242 *10^5 m/s^2

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