A proton with a speed of 2 x 105 m/s falls through a potential difference V and thereby increases its speed to 6 x 105 m/s. Through what potential difference did the proton fall

16.7 x 10²V

The work-energy theorem states that the change in kinetic energy of a particle, Δ, results in work done by the particle, W. i.e;

Δ = W                     ------------------(i)

But:

Δ = mv² -

Δ = m [v² - u²]    ------------------(ii)

Where;

m = mass of particle (proton in this case) = 1.67 x 10⁻²⁷kg

v = final velocity of the particle = 6 x 10⁵m/s

u = initial velocity of the particle = 2 x 10⁵m/s

Substitute these values into equation (ii) as follows;

Δ = (1.67 x 10⁻²⁷) [(6 x 10⁵)² - (2 x 10⁵)²]

Δ = (1.67 x 10⁻²⁷) [(36 x 10¹⁰) - (4 x 10¹⁰)]

Δ = (1.67 x 10⁻²⁷) [(36 - 4) x 10¹⁰]

Δ = (1.67 x 10⁻²⁷) [32 x 10¹⁰]

Δ = (1.67 x 10⁻²⁷) [16 x 10¹⁰]

Δ = 26.72 x 10⁻¹⁷J

Also:

W = qΔV         ----------------(iii)

Where;

q = charge of the particle (proton) = 1.6 x 10⁻¹⁹C

ΔV = change in potential difference of the particle = V (from the question)

Substitute these values into equation (iii) as follows;

W = 1.6 x 10⁻¹⁹ x V        

W = 1.6 x 10⁻¹⁹V         -------------------(iv)

Now:

Substitute the values of Δ = 26.72 x 10⁻¹⁷ and W in equation(iv) into equation (i)

26.72 x 10⁻¹⁷ = 1.6 x 10⁻¹⁹V

Solve for V;

V = (26.72 x 10⁻¹⁷) / (1.6 x 10⁻¹⁹)

V = 16.7 x 10²V

Therefore, the potential difference through which the proton fell was 16.7 x 10²V

a small object of mass 3.66 g and charge -17.6 µc is suspended motionless above the ground when immersed in a uniform electric field perpendicular to the ground. what are the magnitude and direction of the electric field?

1/0.14nm = 10^10/1.4 a lot