Question 24.89: Initially two electrons are fixed in place with a separation......
Initially two electrons are fixed in place with a separation of 2.00 μm. How much work must we do to bring a third electron in from infinity to complete an equilateral triangle?
The net potential at point P (the place where we are to place the third electron) due to the fixed charges is computed using Eq. 24-27 (which assumes V → 0 as r → ∞):
V=\sum_{i=1}^n V_i=\frac{1}{4 \pi \varepsilon_0} \sum_{i=1}^n \frac{q_i}{r_i} (n charged particles). (24-27)
V_P=\frac{-e}{4 \pi \varepsilon_0 d}+\frac{-e}{4 \pi \varepsilon_0 d}=-\frac{2 e}{4 \pi \varepsilon_0 d} .
Thus, with d = 2.00 × 10^{-6} m and e=1.60 \times 10^{-19} \,C , we find
V_P=-\frac{2 e}{4 \pi \varepsilon_0 d}=-\frac{\left(8.99 \times 10^9\, N \cdot m ^2 / C ^2\right)(2)\left(1.60 \times 10^{-19} \, C \right)}{2.00 \times 10^{-6}\, m }=-1.438 \times 10^{-3}\, V .
Then the required “applied” work is, by Eq. 24-14,
\begin{aligned} 1 eV & =e(1 V ) \\ & =\left(1.602 \times 10^{-19} C \right)(1 J / C )=1.602 \times 10^{-19} J \end{aligned} (24-14)
W_{ app }=(-e) V_P=2.30 \times 10^{-22}\, J