Question

speed of an electron between two charges
two stationary positive point - charges, charge 1 of magnitude 3.10 nc and charge 2 of magnitude 1.85 nc, are separated by a distance of 42.0 cm. an electron is released from rest at the point midway between the two charges, and it moves along the line connecting the two charges.
part a
what is the speed $v_{final}$ of the electron when it is 10.0 cm from charge 1?
express your answer in meters per second.
view available hint(s)
$v_{final}=$ m/s

Explanation:

Step1: Calculate initial electric - potential energy

The electric - potential energy of a point charge $q$ in the presence of other charges is given by $U = k\frac{qQ}{r}$. The initial position of the electron is at the mid - point between the two charges. The distance from the electron to each charge at the initial position $r_i=\frac{42.0}{2}=21.0\ cm = 0.21\ m$. The charge of an electron $q=- 1.6\times10^{-19}\ C$, $Q_1 = 3.10\times10^{-9}\ C$ and $Q_2 = 1.85\times10^{-9}\ C$. The initial electric - potential energy $U_i=kq(\frac{Q_1}{r_i}+\frac{Q_2}{r_i})$, where $k = 9\times10^{9}\ N\cdot m^{2}/C^{2}$.
\[

$$\begin{align*} U_i&=9\times10^{9}\times(-1.6\times10^{-19})\times(\frac{3.10\times10^{-9}}{0.21}+\frac{1.85\times10^{-9}}{0.21})\\ &=9\times10^{9}\times(-1.6\times10^{-19})\times\frac{(3.10 + 1.85)\times10^{-9}}{0.21}\\ &=9\times10^{9}\times(-1.6\times10^{-19})\times\frac{4.95\times10^{-9}}{0.21}\\ &=-9\times10^{9}\times1.6\times10^{-19}\times\frac{4.95\times10^{-9}}{0.21}\\ \end{align*}$$

\]
\[U_i\approx - 3.4\times10^{-17}\ J\]

Step2: Calculate final electric - potential energy

The final distance from the electron to charge 1 is $r_{1f}=10.0\ cm = 0.10\ m$, and the distance from the electron to charge 2 is $r_{2f}=42.0 - 10.0=32.0\ cm = 0.32\ m$. The final electric - potential energy $U_f=kq(\frac{Q_1}{r_{1f}}+\frac{Q_2}{r_{2f}})$.
\[

$$\begin{align*} U_f&=9\times10^{9}\times(-1.6\times10^{-19})\times(\frac{3.10\times10^{-9}}{0.10}+\frac{1.85\times10^{-9}}{0.32})\\ &=9\times10^{9}\times(-1.6\times10^{-19})\times(3.10\times10^{-8}+ 5.78\times10^{-9})\\ &=9\times10^{9}\times(-1.6\times10^{-19})\times(3.10\times10^{-8}+0.578\times10^{-8})\\ &=9\times10^{9}\times(-1.6\times10^{-19})\times3.678\times10^{-8}\\ &=-9\times10^{9}\times1.6\times10^{-19}\times3.678\times10^{-8}\\ \end{align*}$$

\]
\[U_f\approx - 5.3\times10^{-17}\ J\]

Step3: Use conservation of energy

By the conservation of energy $\Delta K=-\Delta U$, where $\Delta K = K_f - K_i$ and $K_i = 0$ (released from rest). So $K_f=- (U_f - U_i)$. The kinetic energy $K_f=\frac{1}{2}mv^{2}$, where the mass of an electron $m = 9.11\times10^{-31}\ kg$.
\[

$$\begin{align*} \frac{1}{2}mv^{2}&=-(U_f - U_i)\\ v&=\sqrt{\frac{-2(U_f - U_i)}{m}}\\ &=\sqrt{\frac{-2(-5.3\times10^{-17}+3.4\times10^{-17})}{9.11\times10^{-31}}}\\ &=\sqrt{\frac{-2\times(-1.9\times10^{-17})}{9.11\times10^{-31}}}\\ &=\sqrt{\frac{3.8\times10^{-17}}{9.11\times10^{-31}}}\\ &=\sqrt{4.17\times10^{13}}\\ &\approx6.46\times10^{6}\ m/s \end{align*}$$

\]

Answer:

$6.46\times10^{6}$