A proton moves through an electric potential created by a number of source charges. Its speed is 2.5X105 m/s at a point where the potential is 1500 V. What will be the proton

(6.67 × 10⁵) m/s

Explanation:

What will be the proton's speed a short time later when it reaches a point where the potential is -500V.

According to the law of comservation of energy, the sum of potential and kinetic energy of the proton at any point in time is always constant.

P.E₁ + K.E₁ = P.E₂ + K.E₂

P.E₁ = U₁ = qV₁

q = charge on the proton = (1.602 × 10⁻¹⁹) C

V₁ = 1500 V

P.E₁ = U₁ = (1.602 × 10⁻¹⁹) × (1500) = (2.403 × 10⁻¹⁶) J

K.E₁ = (1/2)mv₁²

m = mass of a proton = (1.673 × 10⁻²⁷) kg

v₁ = (2.5 × 10⁵) m/s

K.E₁ = (1/2)(1.673×10⁻²⁷)(2.5 × 10⁵)² = (5.23 × 10⁻¹⁷) J

P.E₂ = U₂ = qV₂

q = charge on the proton = (1.602 × 10⁻¹⁹) C

V₂ = -500 V

P.E₂ = U₂ = (1.602 × 10⁻¹⁹) × (-500) = (-8.01 × 10⁻¹⁷) J

K.E₂ = ?

P.E₁ + K.E₁ = P.E₂ + K.E₂

(2.403 × 10⁻¹⁶) + (5.23 × 10⁻¹⁷) = (-8.01 × 10⁻¹⁷) + K.E₂

K.E₂ = (2.403 × 10⁻¹⁶) + (5.23 × 10⁻¹⁷) + (8.01 × 10⁻¹⁷) = (3.727 × 10⁻¹⁶) J

K.E₂ = (1/2)mv₂²

m = (1.673 × 10⁻²⁷) kg

v₂ = ?

(3.727 × 10⁻¹⁶) = (1/2)(1.673 × 10⁻²⁷)v₂²

v₂² = (4.455 × 10¹¹)

v₂ = 667,457.863839 m/s = (6.67 × 10⁵) m/s

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