A parallel plate capacitor having cross-sectional area 'A' and separated by distance 'd' is filled by copper plate of thickness b. It's capacitance is :
Understanding Capacitance with a Conductor Plate
Let's analyze how the capacitance of a parallel plate capacitor changes when a copper plate, which is a conductor, is inserted between the plates.
A standard parallel plate capacitor with vacuum or air between the plates has a capacitance given by the formula:
\begin{equation*} C_0 = \frac{\varepsilon_0 A}{d} \end{equation*}
where:
- \(\varepsilon_0\) is the permittivity of free space.
- A is the cross-sectional area of the plates.
- d is the distance between the plates.
Now, consider the situation where a copper plate of thickness 'b' is inserted between the plates. Copper is a conductor. When a conductor is placed in an electric field, charges within the conductor rearrange themselves such that the electric field inside the conductor becomes zero.
Imagine the original parallel plate capacitor with plates at separation 'd'. When a conductor plate of thickness 'b' is inserted parallel to the capacitor plates, the electric field exists only in the space between the capacitor plates and the conductor plate.
The total distance 'd' is now composed of the thickness of the copper plate 'b' and the remaining gap between the capacitor plates, which is \(d - b\). Since the electric field is zero within the copper plate, the potential difference between the capacitor plates is developed only across the regions where the electric field is non-zero, i.e., the gap of thickness \(d - b\).
Effectively, the system behaves like a parallel plate capacitor with a reduced separation equal to the distance where the electric field is present. This effective distance is \(d - b\).
Using the formula for a parallel plate capacitor with the effective distance, the new capacitance \(C\) is:
\begin{equation*} C = \frac{\varepsilon_0 A}{(d - b)} \end{equation*}
This formula shows that inserting a conductor plate between the capacitor plates increases the capacitance because the effective distance over which the electric field acts is reduced.
Let's compare this derived formula with the given options:
- Option 1: \(\frac{\varepsilon_0 A}{2d}\)
- Option 2: \(\frac{\varepsilon_0 A}{d-b}\)
- Option 3: \(\frac{2\varepsilon_0 A}{d+\frac{b}{2}}\)
- Option 4: \(\frac{\varepsilon_0 A}{d+\frac{b}{2}}\)
Our derived capacitance formula is \(\frac{\varepsilon_0 A}{d-b}\), which matches Option 2.
| Parameter | Description | Original Capacitor | Capacitor with Copper Plate |
|---|---|---|---|
| Plate Area | A | A | A |
| Plate Separation | d | d | d |
| Conductor Thickness | b | 0 | b |
| Effective Distance for E field | d | d - b | |
| Capacitance Formula | \(\frac{\varepsilon_0 A}{d}\) | \(\frac{\varepsilon_0 A}{d-b}\) |
Revision Table: Parallel Plate Capacitor Concepts
| Concept | Formula / Key Idea | Notes |
|---|---|---|
| Capacitance Definition | \(C = \frac{Q}{V}\) | Charge per unit potential difference. |
| Parallel Plate Capacitance (Vacuum) | \(C_0 = \frac{\varepsilon_0 A}{d}\) | A = Area, d = Separation. |
| Parallel Plate Capacitance (Dielectric) | \(C = \frac{\kappa \varepsilon_0 A}{d}\) | \(\kappa\) = Dielectric constant. |
| Effect of Conductor Slab | Reduces effective distance for E field to \(d-b\). | E field is zero inside conductor. |
| Capacitance with Conductor Slab | \(C = \frac{\varepsilon_0 A}{d-b}\) | b = thickness of conductor slab. |
Additional Information on Capacitance and Conductors
When a conductor is placed in an external electric field, free charges within the conductor move until the electric field inside the conductor becomes exactly zero. This happens because the induced charges on the surface of the conductor create an internal electric field that opposes the external field.
In the case of a parallel plate capacitor, inserting a conductor slab effectively splits the gap 'd' into three regions: the gap between the positive plate and the conductor (\(d_1\)), the conductor itself (thickness 'b'), and the gap between the conductor and the negative plate (\(d_2\)). The total distance is \(d = d_1 + b + d_2\). The electric field is zero in the conductor (region 'b'). The electric field is present in the gaps \(d_1\) and \(d_2\). The total potential difference V between the plates is the sum of the potential differences across these gaps: \(V = V_1 + V_2\). Since the electric field is uniform (assuming large plates and ignoring edge effects), \(V_1 = E \cdot d_1\) and \(V_2 = E \cdot d_2\), where E is the electric field in the gaps. Thus, \(V = E(d_1 + d_2)\). The total effective distance over which the potential difference is developed is \(d_1 + d_2 = d - b\).
The capacitance is \(C = Q/V\). The electric field in the gap is related to the charge density \(\sigma\) on the plates by \(E = \sigma/\varepsilon_0 = Q/(A\varepsilon_0)\). Substituting this into the potential difference equation, \(V = \frac{Q}{A\varepsilon_0}(d-b)\). Therefore, the capacitance is \(C = \frac{Q}{V} = \frac{Q}{\frac{Q}{A\varepsilon_0}(d-b)} = \frac{A\varepsilon_0}{d-b}\).
This confirms the formula and explains why the thickness of the conductor 'b' is subtracted from the total separation 'd'.
Important Questions from Capacitance
In Maxwell's revision of Ampere's circuital law, the displacement current density, $\vec{J_D}$, was introduced to ensure consistency and is explicitly defined as being directly proportional to:
The unit of capacitance is farad. 1 farad is equal to _________.
Which of the following components store energy in the form of electrical charges?
The capacitance of a capacitor is given by C = Q/V. The capacitance depends on ______.
Whose SI unit is Farad?
