An electron with an initial speed of 460,000 m/s is brought to rest by an electric field. What is the potential difference that stops the electron?

• A. -0.60 V
• B. 0.60 V
• C. -6.0 V
• D. -0.06 V

Answer: (A)

When the electron comes to rest, all its initial kinetic energy must be converted into electric potential energy. The change in potential energy is ΔU = U_f − U_i = K_i, since K_f = 0.

For a charge q in a potential V, U = qV, so ΔU = qΔV. With the electron’s charge q = −e, the potential difference it experiences is ΔV = ΔU/q = K_i/(-e) = −K_i/e.

Compute the initial kinetic energy:

K_i = (1/2) m_e v^2 = (1/2)(9.11×10^−31 kg)(4.60×10^5 m/s)^2 ≈ 9.64×10^−20 J.

Then:

ΔV ≈ −(9.64×10^−20 J)/(1.60×10^−19 C) ≈ −0.60 V.

So the potential difference that stops the electron is about −0.60 V. The negative sign comes from the electron’s negative charge, meaning the potential must drop by ~0.60 V along its path to bring it to rest.