Question

An electron moving parallel to the x-axis has an initial speed of 3.70 x 106 m/s at the origin. Its speed is reduced to 1.40 x 105 m/s at the point x=2.00 cm. (a) Calculate the electric potential difference between the origin and that point. (b) Which point is at the higher potential?

Answer 1

(a) If no nonconservative forces act within a system, the mechanical energy of the system is conserved:

∆K + ∆U = 0

The kinetic energy of a particle of mass m moving with a speed v is defined as:

$K = \frac{1}{2}mv^2$

The potential difference ∆V between two points is the change in potential energy ∆U per unit charge:

$∆V = \frac{∆U}{q}$

$(K_f – K_i) + ∆U = 0$

$(\frac{1}{2}m_ev_f^2 - \frac{1}{2}m_ev_i^2) + q_e∆V = 0$

$\frac{1}{2}m_e(v_f^2 - v_i^2) = -q_e∆V$

$∆V = -\frac{1}{2q}m_e(v_f^2 - v_i^2)$

$∆V = -\frac{1}{2(-1.602 \times 10^{-19}}(9.11 \times 10^{-31})[(1.4 \times 10^5)^2 – (3.7 \times 10^6)^2] = -38.9 \;V$

(b) Because the potential difference ∆V is negative, the initial electric potential is greater than the final potential energy. Therefore, the potential at the origin will be higher.