Question 24.92: In Fig. 24-70, point P is at the center of the rectangle. Wi......
In Fig. 24-70, point P is at the center of the rectangle. With V = 0 at infinity, q_1 = 5.00 fC, q_2 = 2.00 fC, q_3 = 3.00 fC, and d = 2.54 cm, what is the net electric potential at P due to the six charged particles?
The net electric potential at point P is the sum of those due to the six charges:
\begin{aligned} V_P= & \sum_{i=1}^6 V_{Pi}=\sum_{i=1}^6 \frac{q_i}{4 \pi \varepsilon_0 r_i}=\frac{10^{-15}}{4 \pi \varepsilon_0}\left[\frac{5.00}{\sqrt{d^2+(d / 2)^2}}+\frac{-2.00}{d / 2}+\frac{-3.00}{\sqrt{d^2+(d / 2)^2}}\right. \\ & \left.+\frac{3.00}{\sqrt{d^2+(d / 2)^2}}+\frac{-2.00}{d / 2}+\frac{+5.00}{\sqrt{d^2+(d / 2)^2}}\right]=\frac{9.4 \times 10^{-16}}{4 \pi \varepsilon_0\left(2.54 \times 10^{-2}\right)} \\ = & 3.34 \times 10^{-4} \,V . \end{aligned}