Question

A parallel-plate capacitor consists of two plates, each with an area of 25cm2 separated by 3.0 mm. The charge on the capacitor is 9.3nC . A proton is released from rest next to the positive plate. How long does it take for the proton to reach the negative plate?

Answer 1

Electric Field between capacitor plates = Q/(A*E₀)

Q = 9.3*10^-9 C

A = 25*10^-4 m²

E₀ = 8.854*10^-12 F/m

Electric Field = (9.3*10^-9)/(25*10^-4*8.854*10^-12) = 4.2015*10⁵ N/C

Force on Proton = (1.602 x 10^-19)*(4.2015*10⁵) = 6.7224*10^-14 N

Acceleration of proton = F/m = (6.7224*10^-14)/(1.67262178 10^-27) = 4.019*10¹³ m/s²

Now, v² = 2*a*s = 2* 4.019*10¹³*3*10^-3

v = 4.91068*10⁵ m/s

Now, v = a*t

t = v/a = (4.91068*10⁵)/(4.019*10¹³) = 1.2218664*10^-8 sec

Time = 1.222*10^-8 seconds

Answer 2

Note that the electric field between capacitors is

E = Q/ [Eo A]

Thus, as Q = 5.3E-9 C, A = 0.0025 m^2,

E = 2.395E5 N/C

Thus, the force on the proton is, for a charge q = 1.602E-19 C,

F = Eq = 3.8376E-14 N

Thus, as its mass is 1.67E-27 kg, a = F/m,

a = 2.2979E13 m/s^2

Thus, as

d = vo t + 1/2 a t^2

Here, vo = 0, d = 0.003 m, then

t = sqrt(2d/a)

= 1.62*10^-8 s   [ANSWER]

Answer 3

SInce C = Q/V we can find the potential difference, V. First, C is computed using

C = e0*A/d where e0 = 8.85x10^-12 C^2/N-m^2 and A = 25 cm^2 = 2.5x10^-3 m^2 d = 3x10^-3 m

C = 7.37x10^-12 F = 7.37 pF

Then V = Q/C = 9.8x10^-9C/7.37x10^-12 F = 1329 Volts

Lets use conservation of energy to find the speed the proton that reaches the other plate:

U = eV = 1/2 mv^2 --> v = sqrt(2eV/m) = sqrt(2*1.6*10^-19*1329/9.1*10^-31) = 21.61*10^6 m/s

Then using d = 1/2 at^2 , we write

t = 2d/v = 2*0.003m/21.61*10^6 m/s = 2.77 x10^-10 sec