---
title: "Copper slab inserted in capacitor"
subject: "Engineering Physics"
chapter: 7
key: "semester_1/engineering_physics:copper-slab-capacitor-capacitance-change"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/copper-slab-capacitor-capacitance-change
site: BCT Engineering
---

# Copper slab inserted in capacitor

Engineering Physics · Chapter 7: Electrostatics

- Asked in: 2067 Asar, 5 marks
- Video: [Capacitors - Inserting a metal slab between the plates](https://www.youtube.com/watch?v=AzQcGf0KHaE) — Physics Ninja
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/engineering_physics/copper-slab-capacitor-capacitance-change
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-7

## Question

A copper slab of thickness b is inserted into a parallel plate capacitor exactly half way between the plates. If the separation of the plates is d and the area of each plate is A, show that the change in capacitance is equal to ε0Ab/((d − b)d).

## Short answer

Initially C₀ = ε₀A/d. With the slab, only air gaps of total thickness d−b sustain field, so C = ε₀A/(d−b). The change ΔC = ε₀A b / [(d−b)d].

## Answer

Parallel plate capacitor: plate area A, separation d. Without the slab, C₀ = ε0A/d.

```text
   ______________________  + plate
        d1 (air)
   ======================  copper slab, thickness b
        d2 (air)
   ______________________  - plate
   d1 + d2 = d - b ; here d1 = d2 = (d-b)/2
```

#### With the slab
1. Charge +q on the top plate induces −q on the upper face of the slab and +q on the lower face. The field inside copper is zero.
2. Field exists only in the air gaps: E = q/(ε0 A) in each gap (Gauss's law).
3. Potential difference between the plates = field × total air thickness:

V = E (d₁ + d₂) = (q/ε0A)(d − b)

4. New capacitance:

C = q/V = ε0A / (d − b)

The result depends only on the total air gap (d − b), not on where the slab sits. The two gaps behave as two capacitors in series: 1/C = d₁/ε0A + d₂/ε0A = (d − b)/ε0A.

#### Change in capacitance
ΔC = C − C₀ = ε0A/(d − b) − ε0A/d = ε0A [ d − (d − b) ] / [ d (d − b) ]

**ΔC = ε0 A b / [ (d − b) d ]**

Shown. The capacitance increases, since the effective separation shrinks from d to d − b.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What is the field inside the copper slab?
   A. E = q/(ε0 A)
   B. Half the gap field
   C. Zero
   D. Equal to V/d
   Answer: C. Conductors have no electrostatic field inside, so only the air gaps carry field. (Key point: Field inside copper)

2. The potential difference across the plates is
   A. (q/ε0A) d
   B. (q/ε0A)(d − b)
   C. (q/ε0A) b
   D. (q/ε0A)(d + b)
   Answer: B. The field exists only in the air gaps of total thickness d − b. (Key point: Potential difference)

3. How does the new capacitance depend on slab position?
   A. It is highest at the centre
   B. It is highest next to a plate
   C. It depends only on d − b, not on position
   D. It is unchanged by the slab
   Answer: C. The two air gaps add in series: 1/C = (d1 + d2)/ε0A = (d − b)/ε0A. (Key point: Position independence)

4. ΔC = C − C0 equals
   A. ε0A b/((d−b)d)
   B. ε0A/(d−b)
   C. ε0A b/d^2
   D. −ε0A b/((d−b)d)
   Answer: A. ε0A/(d−b) − ε0A/d = ε0A b/(d(d−b)), an increase. (Key point: Change)

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