A proton is released from rest in a uniform E – Field 8 x 104V/m directed alomg + ve x – axis, undergoes a displacement of 0. 5m in direction of field. Find: a) Change of potential for this displacement (A =» B)? B) Change of P. E for this displacement (A =» B)? C) The speed of proton after the displacement?

a. -40 kV b. -6.408 fJ c. 2.77 × 10⁶ m/s

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

where are the vectors ?

spare some change and give it to the poor

explanation:

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