Use the parallel-plate capacitor shown below; V = 0 X = 0 The boundary conditions are indicated on the figure as; V = 0 at X = 0 V=V X = X V = V at X = X The potential, V between the plates is a linear function of the spacing, x between the plates given by V = Ax Where x is the separation between the plates and A is a constant, determined by the boundary conditions. i. Show that for V = Ax to be a unique solution for the potential variation between the plates, A has to be; V A = ii. If A was now changed to A = Bx (3 marks) Show that the solution in this case, for the voltage V = Ax will not be a unique solution for the voltage variation. (4 marks)