Question 3.18: Find the values of the charges that will cause the potential...
Find the values of the charges that will cause the potentials as shown in the Figure 3-18, if the coordinates of the points in the plane z = 0 are: Q1 (2,3) , Q2 (2,2) , Q3 (5,3) and Q4 (5,2) . Assume that the diameter of the charges is 2a = 1 meter.
The matrix [P] in (3.83) and (3.84) has the elements
[ P ]=\frac{1}{4 \pi \varepsilon_{0}}\left[\begin{array}{cccc} \frac{1}{\left| r _{1}- r _{1}\right|} & \frac{1}{\left| r _{1}- r _{2}\right|} & \frac{1}{\left| r _{1}- r _{3}\right|} & \frac{1}{\left| r _{1}- r _{4}\right|} \\ \frac{1}{\left| r _{2}- r _{1}\right|} & \frac{1}{\left| r _{2}- r _{2}\right|} & \frac{1}{\left| r _{2}- r _{3}\right|} & \frac{1}{\left| r _{2}- r _{4}\right|} \\ \frac{1}{\left| r _{3}- r _{1}\right|} & \frac{1}{\left| r _{3}- r _{2}\right|} & \frac{1}{\left| r _{3}- r _{3}\right|} & \frac{1}{\left| r _{3}- r _{4}\right|} \\ \frac{1}{\left| r _{4}- r _{1}\right|} & \frac{1}{\left| r _{4}- r _{2}\right|} & \frac{1}{\left| r _{4}- r _{3}\right|} & \frac{1}{\left| r _{4}- r _{4}\right|} \end{array}\right] (3.83)
P_{i, i}=\frac{1}{4 \pi \varepsilon_{0} a} (3.84)
[P]=\frac{1}{4 \pi \varepsilon_{0}}\left[\begin{array}{lllll} \frac{1}{(1 / 2)} & \frac{1}{1} & \frac{1}{3} & \frac{1}{\sqrt{1^{2}+3^{2}}} \\ \frac{1}{1} & \frac{1}{(1 / 2} & \frac{1}{\sqrt{1^{2}+3^{2}}} & \frac{1}{3} \\ \frac{1}{3} & \frac{1}{\sqrt{1^{2}+3^{2}}} & \frac{1}{(1 / 2)} & \frac{1}{1} \\ \frac{1}{\sqrt{1^{2}+3^{2}}} & \frac{1}{3} & \frac{1}{1} & \frac{1}{(1 / 2)} \end{array}\right]=\frac{1}{4 \pi \varepsilon_{0}}\left[\begin{array}{ccccc} 2 & 1 & 0.3333 & 0.3162 \\ 1 & 2 & 0.3162 & 0.3333 \\ 0.3333 & 0.3162 & 2 & 1 \\ 0.3162 & 0.3333 & 1 & 2 \end{array}\right] .
The column vector for the potential is [ V ]=[-1,-1,1,1]^{ T}where the “T” indicates the transpose. Solving the matrix equation (3.82) leads to
[P][Q] = [V] (3.82)
Q _{1}= Q _{2}=- Q _{3}=- Q _{4}=-.4254(4 \pi \varepsilon_{0})