Chapter 24: Q79P (page 715)

An electron is released from rest on the axis of an electric dipole that has charge eand charge separation $d = 20 pm$ pm and that is fixed in place. The release point is on the positive side of the dipole, at distance $7.0d$ from the dipole center. What is the electron’s speed when it reaches a point $5.0d$ from the dipole center?

Short Answer

Expert verified

$v = 7.0 \times 10^{5} m/s$

Step by step solution

Given

$d = 20 pm = 20 \times 10^{- 12} m$

Understanding the concept

After reading the question, The electric potential energy in the presence of a dipole is $U = {qV}_{dipole}$ .

$\begin{matrix} U = \frac{qp cos θ}{4 {πε}_{o} r^{2}} \\ U = \frac{(- e) (ed) cos θ}{4 {πε}_{o} r^{2}} \\ θ_{i} = θ_{f} = 0^{\circ} . \end{matrix}$

Apply the Law of Conservation of Energy: $K_{f} + U_{f} = K_{i} + U_{i}$ .

Calculate the electron’s speed when it reaches a point 5.0d from the dipole center

$K_{f} + U_{f} = K_{i} + U_{i}$

$\begin{matrix} \frac{1}{2} {mv}^{2} = \frac{e^{2} d}{4 {πε}_{o} d^{2}} (\frac{1}{5^{2}} - \frac{1}{7^{2}}) \\ = \frac{e^{2}}{4 {πε}_{o} d} (\frac{1}{25} - \frac{1}{49}) \\ v^{2} = \frac{2 e^{2}}{4 {πε}_{o} dm} (\frac{1}{25} - \frac{1}{49}) d = 20 \times 10^{- 12} m, m = 9.11 \times 10^{- 31} kg \\ v = \sqrt{\frac{2 e^{2}}{4 {πε}_{o} dm} (\frac{1}{25} - \frac{1}{49})} \\ v = 7.0 \times 10^{5} m/s \end{matrix}$