Question

Two parallel plates having charges of equal magnitude but opposite sign are separated by 29.0 cm. Each plate has a surface charge density of 32.0 nC/m2. A proton is released from rest at the positive plate. (a) Determine the magnitude of the electric field between the plates from the charge density. kN/C (b) Determine the potential difference between the plates. V (c) Determine the kinetic energy of the proton when it reaches the negative plate. J (d) Determine the speed of the proton just before it strikes the negative plate. km/s (e) Determine the acceleration of the proton. Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. m/s2 towards the negative plate (f) Determine the force on the proton. N towards the negative plate (g) From the force, find the magnitude of the electric field. kN/C (h) How does your value of the electric field compare with that found in part (a)

Answer 1

Answer:

(a) E = 3.6 x 10³ N/C = 3.6 KN/C

(b) ΔV = 1044 Volts

(c) K.E = 1.67 x 10⁻¹⁶ J

(d) Vf = 4.47 x 10⁵ m/s

(e) a = 3.45 x 10¹¹ m/s²

(f) F = 5.76 x 10⁻¹⁶ N

(g) E = 3.6 x 10³ N/C = 3.6 KN/C  

(h)  Both values are same in part (h) and (a)

Explanation:

(a)

Electric field between oppositely charged plates is given as follows:

E = σ/ε₀

where,

E = Electric Field Intensity = ?

σ = surface charge density = 32 nC/m² = 3.2 x 10⁻⁸ C/m²

ε₀ = Permittivity of free space = 8.85 x 10⁻¹² C²/N.m²

Therefore,

E = (3.2 x 10⁻⁸ C/m²)/(8.85 x 10⁻¹² C²/N.m²)

E = 3.6 x 10³ N/C = 3.6 KN/C

(b)

E = ΔV/r

ΔV = Er

where,

r = distance between plates = 29 cm = 0.29 m

ΔV = Potential Difference = ?

ΔV = (3.6 x 10³)(0.29)

V = 1044 Volts

(c)

Kinetic Energy of Proton = Work done on Proton

K.E = F r

but,  F = E q

K.E = E q r

where,

q = charge on proton = 1.6 x 10⁻¹⁹ C

Therefore,

K.E = (3600 N/C)(1.6 x 10⁻¹⁹ C)(0.29 m)

K.E = 1.67 x 10⁻¹⁶ J

(d)

K.E = (1/2)m(Vf² - Vi²)

where,

m = mass of proton = 1.67 x 10⁻²⁷ kg

Vf = Final Velocity = ?

Vi = Initial Velocity = 0 m/s

Therefore,

1.67 x 10⁻¹⁶ J = (1/2)(1.67 x 10⁻²⁷ kg)(Vf² - (0 m/s)²]

Vf² = (1.67 x 10⁻¹⁶ J)(2)/(1.67 x 10⁻²⁷ kg)

Vf = √(20 x 10¹⁰ m²/s²)

Vf = 4.47 x 10⁵ m/s

(e)

2as = Vf² - Vi²

2(a)(0.29 m) = (4.47 x 10⁵ m/s)² - (0 m/s)²

a = (20 x 10¹⁰ m²/s²)/0.58 m

a = 3.45 x 10¹¹ m/s²

(f)

F = ma

F = (1.67 x 10⁻²⁷ kg)(3.45 x 10¹¹ m/s²)

F = 5.76 x 10⁻¹⁶ N

(g)

E = F/q

E = (5.76 x 10⁻¹⁶ N)/(1.6 x 10⁻¹⁹ C)

E = 3.6 x 10³ N/C = 3.6 KN/C

(h)

Both values are same in part (h) and (a)