Potential energy of point-charges: If an electron is accelerated from rest through a potential difference of 9.9 kV, what is its resulting speed? (e = 1.60 × 10^-19 C, k = 1/4πε₀ = 8.99 × 10⁹ N ∙ m²/C²,^m_el= 9.11 × 10^-31 kg)

Question

Quizplus · Accepted Answer

We can use the formula for potential energy of a point charge:

U = keqQ/r

where ke is Coulomb's constant (k), q is the charge of the electron, Q is the charge creating the potential difference, and r is the distance between the charges.

Since the electron is accelerated through a potential difference of 9.9 kV, we can find the change in potential energy:

ΔU = qΔV

where ΔV is the potential difference.

ΔU = (1.60 × 10^-19 C)(9.9 × 10^3 V) = 1.584 × 10^-15 J

To find the resulting speed, we can equate the change in potential energy to the kinetic energy:

ΔU = K = 1/2mv^2

where m is the mass of the electron and v is its velocity.

Solving for v, we get:

v = sqrt((2ΔU)/m)

Plugging in the values:

v = sqrt((2(1.584 × 10^-15 J))/(9.11 × 10^-31 kg)) ≈ 5.9 × 10^7 m/s ≈ 5.9 × 10^17 cm/s

Converting to SI units:

v ≈ 5.9 × 10^5 m/s

So the best choice is A.

Potential Energy of Point-Charges: If an Electron Is Accelerated from Rest