The electric potential at the midpoint:

$V = k (\frac{q_{1}}{r} + \frac{q_{2}}{r})$

r = d/2 = 46.0/2 = 23.0 cm

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$\begin{array}{rcl} V & = & (9 \times 10^{9}) (\frac{3.90 \times 10^{- 9}}{23.0 \times 10^{- 2}} + \frac{1.80 \times 10^{- 9}}{23.0 \times 10^{- 2}}) \\ = & (9 \times 10^{9}) (1.6957 \times 10^{- 8} + 7 . 826^{- 9}) \end{array}$

V = 223 V

Potential at between the electron and q₁ is:

$\begin{array}{rcl} V_{{eq}_{1}} & = & (9 \times 10^{9}) (\frac{3.90 \times 10^{- 9}}{10 \times 10^{- 2}} + \frac{1.80 \times 10^{- 9}}{(46.0 - 10.0) \times 10^{- 2}}) \\ = & (9 \times 10^{9}) (3.9 \times 10^{- 8} + 5 . 0^{- 9}) \end{array}$

V_eq1 = 396 V

From conservation of energy;

$\begin{array}{rcl} P E_{0} + K E_{0} & = & P E_{f} + K E_{f} \\ \frac{1}{2} m v^{2} & = & ∆ V q_{e} \\ v & = & \sqrt{\frac{2 ∆ V q_{e}}{m}} \\ = & \sqrt{\frac{(2) (396 - 223) (1.6 \times 10^{- 19})}{9.1 \times 10^{- 31}}} \end{array}$

v = 7.8 × 10⁶ m/s

The speed of the electron when it is 10.0 cm from the charge 1 is 7.8 × 10⁶ m/s.

Problem: a. Which point, A or B, has a larger electric potentialb. What is the potential difference between A and B? Assume that E=1500V/m

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