x = 0

and the second at

x = d = x_{o}

ρ

\frac{d ^{2} V}{d x ^{2}} = - \frac{ρ}{ϵ _{0}}, V (0) = V_{1}, V (d) = V_{2} .

Integrating twice and imposing both values yields

V (x) = V_{1} + \frac{V _{2} - V _{1}}{d} x + \frac{ρ}{2 ϵ _{0}} x (d - x) .

E_{x} (x) = \frac{V _{1} - V _{2}}{d} + \frac{ρ}{ϵ _{0}} (x - \frac{d}{2}), ∣ E (x) ∣ = \frac{V _{1} - V _{2}}{d} + \frac{ρ}{ϵ _{0}} (x - \frac{d}{2}) .

Thus the electric potential is a parabola, concave downward for

ρ > 0

and upward for

ρ < 0

. The signed electric field is a straight line. Its magnitude has a V-shaped corner if its zero

x_{*} = \frac{d}{2} - \frac{ϵ _{0} ( V _{1} - V _{2} )}{ρ d}

lies inside the gap; otherwise its magnitude is a single straight segment. An interior zero requires

∣ V_{1} - V_{2} ∣ < ∣ ρ ∣ d^{2} / (2 ϵ_{0})

. At that point the electric potential has its interior extremum. For

ρ = 0

, these reduce to a linear electric potential and a uniform electric field.

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