Q79P An electron is released from res... [FREE SOLUTION]

91影视

Chapter 24: Q79P (page 715)

An electron is released from rest on the axis of an electric dipole that has charge eand charge separation $d = 20 鈥塸尘$ 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鈥檚 speed when it reaches a point $5.0d$ from the dipole center?

Short Answer

Expert verified

$v = 7.0 脳 10^{5} 鈥尘/蝉$

Step by step solution

Given

$d = 20 鈥塸尘 = 20 脳 10^{鈭� 12} 鈥尘$

Understanding the concept

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

$\begin{matrix} U = \frac{qp 鈥塩辞蝉鈥� 胃}{4 {蟺蔚}_{o} r^{2}} \\ U = \frac{(鈭� e) (ed) 肠辞蝉鈥� 胃}{4 {蟺蔚}_{o} r^{2}} \\ 胃_{i} = 胃_{f} = 0^{鈭�} . \end{matrix}$

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

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

$\begin{matrix} \frac{1}{2} {mv}^{2} = \frac{e^{2} d}{4 {蟺蔚}_{o} d^{2}} (\frac{1}{5^{2}} 鈭� \frac{1}{7^{2}}) \\ = \frac{e^{2}}{4 {蟺蔚}_{o} d} (\frac{1}{25} 鈭� \frac{1}{49}) \\ v^{2} = \frac{2 e^{2}}{4 {蟺蔚}_{o} dm} (\frac{1}{25} 鈭� \frac{1}{49}) 鈥夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌� d = 20 脳 10^{鈭� 12} 鈥尘, 鈥� m = 9.11 脳 10^{鈭� 31} 鈥塳驳 \\ v = \sqrt{\frac{2 e^{2}}{4 {蟺蔚}_{o} dm} (\frac{1}{25} 鈭� \frac{1}{49})} \\ v = 7.0 脳 10^{5} 鈥尘/蝉 \end{matrix}$