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)
<u>E = 3.6 x 10³ N/C = 3.6 KN/C</u>
<u></u>
(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)
<u>V = 1044 Volts</u>
<u></u>
(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)
<u>K.E = 1.67 x 10⁻¹⁶ J</u>
<u></u>
(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)
<u>Vf = 4.47 x 10⁵ m/s</u>
<u></u>
(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
<u>a = 3.45 x 10¹¹ m/s²</u>
<u></u>
(f)
F = ma
F = (1.67 x 10⁻²⁷ kg)(3.45 x 10¹¹ m/s²)
<u>F = 5.76 x 10⁻¹⁶ N</u>
<u></u>
(g)
E = F/q
E = (5.76 x 10⁻¹⁶ N)/(1.6 x 10⁻¹⁹ C)
<u>E = 3.6 x 10³ N/C = 3.6 KN/C</u>
<u></u>
(h)
<u>Both values are same in part (h) and (a)</u>
