Q · 23

Copper slab inserted between the plates

Capacitance

Question

A slab of copper of thickness \(b\) is inserted in between the plates of parallel plate capacitor as shown in figure. The separation of the plate is \(d\). If \(b = d/2\), then the ratio of capacities of the capacitor after and before inserting the slab will be

The Arrangement

Cu d b

Copper is a conductor: the field inside it is zero, so it behaves like an extension of the wire. The effective gap shrinks from \(d\) to \(d-b\) — and the slab’s position in the gap doesn’t matter.

Given

Plate separation\(d\)
Slab materialcopper (conductor)
Slab thickness\(b = d/2\)

Asked

Wanted\(\dfrac{C_{\text{after}}}{C_{\text{before}}}\)
Key ideaconductor ⇒ subtract \(b\) from gap

Concept

Inserting a conducting slab of thickness \(b\) leaves only \((d-b)\) of real gap to hold the field. The capacitance becomes

$$ C_{\text{after}} = \frac{\varepsilon_0 A}{\,d-b\,}. $$

This is the conductor rule — independent of \(b\)’s position. A dielectric slab would instead divide by \(\kappa\), not just remove its thickness.

Method & Steps

  1. Before: \(C_{\text{before}} = \dfrac{\varepsilon_0 A}{d}.\)
  2. After: \(C_{\text{after}} = \dfrac{\varepsilon_0 A}{d-b} = \dfrac{\varepsilon_0 A}{d-d/2}.\)
  3. Simplify: \(C_{\text{after}} = \dfrac{\varepsilon_0 A}{d/2} = \dfrac{2\varepsilon_0 A}{d} = 2\,C_{\text{before}}.\)
  4. Ratio: \(\dfrac{C_{\text{after}}}{C_{\text{before}}} = 2:1.\)

Easy Trick

Conductor inside → just subtract its thickness from the gap: new gap \(= d-b\). With \(b=d/2\) the gap halves, so capacitance doubles → \(2:1\). Wherever you slide the copper, the answer is the same.

b
\(\dfrac{C_{\text{after}}}{C_{\text{before}}} = 2:1\)
Copper removes half the gap, doubling the capacitance.