Answer
At $x = 7.0~cm,$ the speed of the electron is $~~8.4\times 10^5~m/s$
Work Step by Step
We can find the total energy at $x = 4.5~cm$:
$E = K+U$
$E = 3.00~eV+(-1e)(6.0~V)$
$E = -3.0~eV$
By conservation of energy, the electron can reach an electric potential as low as $3.0~V$
Since the right side of the region has an electric potential of $5.0~V$, the electron can escape out the right side of the region.
We can find the kinetic energy as it escapes:
$K+U = -3.0~eV$
$K = -3.0~eV-U$
$K = -3.0~eV-(-1e)(5.0~V)$
$K = 2.0~eV$
$K = (2.0~eV)(\frac{1.6\times 10^{-19}~J}{1~eV})$
$K = 3.2\times 10^{-19}~J$
We can find the speed of the electron at $x = 7.0~cm$:
$K = 3.2\times 10^{-19}~J$
$\frac{1}{2}mv^2 = 3.2\times 10^{-19}~J$
$v^2 = \frac{(2)(3.2\times 10^{-19}~J)}{m}$
$v = \sqrt{\frac{(2)(3.2\times 10^{-19}~J)}{m}}$
$v = \sqrt{\frac{(2)(3.2\times 10^{-19}~J)}{9.109\times 10^{-31}~kg}}$
$v = 8.4\times 10^5~m/s$
At $x = 7.0~cm,$ the speed of the electron is $~~8.4\times 10^5~m/s$
