Discussion
Two students solved the Poisson's equation \(\nabla^{2}\varphi = -\frac{\rho_{0}}{\varepsilon_{0}}\) (the symbols have usual meanings). They obtained two different answers \(\varphi_{1}(x) = -\frac{1}{2}\frac{\rho_{0}x^{2}}{\varepsilon_{0}}\) and \(\varphi_{2}(y) = -\frac{1}{2}\frac{\rho_{0}y^{2}}{\varepsilon_{0}}\). What is the reason ?
Answer: C
On the key, the reason is that the boundary conditions are different in the two cases.
Both functions satisfy Poisson's equation: each has \(\nabla^{2}\varphi = -\rho_{0}/\varepsilon_{0}\). Their difference
\[ \begin{aligned} &\varphi_{1} - \varphi_{2} \\ &= -\frac{\rho_{0}}{2\varepsilon_{0}}(x^{2} - y^{2}) \end{aligned} \]
is a solution of Laplace's equation. By the uniqueness theorem, Poisson's equation has exactly one solution once the boundary conditions are fixed, so two different solutions can only arise if the two students imposed different boundary conditions.
Neither answer is wrong, so 'one of them was wrong' fails. Isotropy of space does not make two different functions equal. Adding a Laplace solution describes how the answers differ, but what allows that difference is the change in boundary conditions, which is the reason the key accepts.
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