a proton travelling along the x-axis is slowed by a uniform electric field E. at x = 20 cm, the proton has a speed of 3.5x10^6 m/s and at 80cn the speed is zero. Determine the magnitude and the direction of E

The magnitude of the electric field is 3.8 × 10⁵⁸ N/C and it is in the negative x- direction.

Explanation:

From work-kinetic energy principles, the kinetic energy change of the proton equals the work done by the electric field.

So, ΔK = W

1/2m(v₂² - v₁²) = qEd where m = mass of proton = 1.673 × 10⁻²⁷ kg, v₁ = initial speed of proton = 3.5 × 10⁶ m/s, v₂ = final speed of proton = 0 m/s, q = proton charge = + e = 1.602 × 10⁻¹⁹ C, E = electric field and d = distance moved by proton = x₂ - x₁ , x₁ = 20 cm and x₂ = 80 cm. So, d = 80 cm - 20 cm = 60 cm = 0.6 m

1/2m(v₂² - v₁²) = qEd

E = (v₂² - v₁²)/2mqd

substituting the values of the variables into E, we have

E = ((0 m/s)² - (3.5 × 10⁶ m/s)²)/(2 × 1.673 × 10⁻²⁷ kg × 1.602 × 10⁻¹⁹ C × 0.6 m)

E = - 12.25 × 10¹² m²/s² ÷ 3.22 × 10⁻⁴⁶

E = -3.8 × 10⁵⁸ N/C

So, the magnitude of the electric field is 3.8 × 10⁵⁸ N/C and it is in the negative x- direction.

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