Three conducting plates, each of area A, are connected as shown.

(a) Are the two capacitos formed connected in series or in parallel?

(b) Determine capacitance C as a function of d1, d2, and A. Assume d1 + d2 is much less than the dimensions of the plates.

You have effectively got two capacitors in parallel. The effective capacitance is just the sum of the two.

Cequiv = ε₀A/d₁ + ε₀A/d₂ Take these over a common denominator (d₁d₂)

Cequiv = ε₀d₂A + ε₀d₁A / (d₁d₂) Cequiv = ε₀A ((d₁ + d₂) / (d₁d₂))

B) It's tempting to just wave your arms and say that when d₁ or d₂ tends to zero C - > ∞, so the minimum will occur in the middle, where d₁ = d₂

But I suppose we ought to kick that idea around a bit.

(d₁ + d₂) is effectively a constant. It's the distance between the two outer plates. Call it D.

C = ε₀AD / d₁d₂ We can also say: d₂ = D - d₁ C = ε₀AD / d₁ (D - d₁) C = ε₀AD / d₁D - d₁²

Differentiate with respect to d₁

dC/dd₁ = - ε₀AD (D - 2d₁) / (d₁D - d₁²) ² {d2C/dd₁² is positive so it will give us a minimum} For max or min equate to zero.

-ε₀AD (D - 2d₁) / (d₁D - d₁²) ² = 0 - ε₀AD (D - 2d₁) = 0 ε₀, A, and D are all non-zero, so (D - 2d₁) = 0 d₁ = ½D

In other words when the middle plate is halfway between the two outer plates, (quelle surprise) so that

d₁ = d₂ = ½D so

Cmin = ε₀AD / (½D) ² Cmin = 4ε₀A / D Cmin = 4ε₀A / (d₁ + d₂)