Answer:
a. -40 kV b. -6.408 fJ c. 2.77 × 10⁶ m/s
Step-by-step explanation:
a) Change of potential for this displacement (A =» B)?
The potential energy change ΔV = -EΔx where E = electric field strength = 8 × 10⁴ V/m and Δx = displacement of charge = 0.5 m
So, ΔV = -EΔx
ΔV = -8 × 10⁴ V/m × 0.5 m
ΔV = -4 × 10⁴ V
ΔV = -40 × 10³ V
ΔV = -40 kV
B) Change of P.E for this displacement (A =» B)?
The change in potential energy ΔU = qΔV where q = proton charge = +1.602 × 10⁻¹⁹ C and ΔV = change in potential = -4 × 10⁴ V
So, ΔU = qΔV
ΔU = +1.602 × 10⁻¹⁹ C × -4 × 10⁴ V
ΔU = -6.408 × 10⁻¹⁵ J
ΔU = -6.408 fJ
C) The speed of proton after the displacement?
From the conservation of energy,
The electric potential energy change of the charge equals its kinetic energy change
ΔK + ΔU = 0
ΔK = -ΔU
ΔK = -(-6.408 × 10⁻¹⁵ J)
K₂ - K₁ = 6.408 × 10⁻¹⁵ J
where K₁ = initial kinetic energy of proton = 0 J (since it starts from rest)
K₂ = final kinetic energy of proton = 1/2mv² where m = mass of proton = 1.673 × 10⁻²⁷ kg and v = speed of proton
K₂ - K₁ = 6.408 × 10⁻¹⁵ J
1/2mv² - 0 = 6.408 × 10⁻¹⁵ J
1/2mv² = 6.408 × 10⁻¹⁵ J
v² = 2(6.408 × 10⁻¹⁵ J)/m
v = √[12.816 × 10⁻¹⁵ J)/1.673 × 10⁻²⁷ kg]
v = √[7.66 × 10¹² J/kg]
v = 2.77 × 10⁶ m/s