A parallel-plate capacitor of capacitance C 1is made using two gold plates. Another parallel-plate capacitor of capacitance C 2is made using two aluminium plates with same plate separation, and all the four plates are of same area. If ρ gand ρ aare respectively the electrical resistivity of gold and aluminium, then which one of the following relations is correct?
C 1= C 2
Parallel Plate Capacitor Capacitance Relation Explained
The question asks us to determine the correct relationship between the capacitance of two parallel-plate capacitors, C1 and C2. Both capacitors have the same plate area and the same plate separation. Capacitor C1 uses gold plates, and capacitor C2 uses aluminium plates. We are given the electrical resistivities of gold (\(\rho_g\)) and aluminium (\(\rho_a\)).
The capacitance (\(C\)) of a parallel-plate capacitor is given by the formula:
\(C = \frac{\epsilon A}{d}\)
Where:
- \(C\) is the capacitance
- \(\epsilon\) is the permittivity of the dielectric material between the plates
- \(A\) is the area of one of the plates
- \(d\) is the separation between the plates
Let's analyze the factors given in the question for both capacitors:
For capacitor C1 (gold plates):
- Plate material: Gold
- Plate area: \(A\) (same as C2)
- Plate separation: \(d\) (same as C2)
- Permittivity of dielectric: Let's assume the space between the plates is filled with the same dielectric material (e.g., vacuum or air) for both capacitors. Let this permittivity be \(\epsilon\).
So, the capacitance of C1 is \(C_1 = \frac{\epsilon A}{d}\).
For capacitor C2 (aluminium plates):
- Plate material: Aluminium
- Plate area: \(A\) (same as C1)
- Plate separation: \(d\) (same as C1)
- Permittivity of dielectric: Same as C1, which is \(\epsilon\).
So, the capacitance of C2 is \(C_2 = \frac{\epsilon A}{d}\).
Factors Affecting Parallel Plate Capacitor Capacitance
From the formula \(C = \frac{\epsilon A}{d}\), we can see that the capacitance of a parallel-plate capacitor depends on the following factors:
- The permittivity (\(\epsilon\)) of the dielectric material between the plates.
- The area (\(A\)) of the plates.
- The distance (\(d\)) between the plates.
The material of the plates themselves (gold or aluminium in this case) is a conductor. While ideal conductors are assumed in the basic formula, the capacitance is determined by the geometry (A and d) and the properties of the *insulating* material (dielectric) between the plates, which facilitates the storage of electrical energy by supporting the electric field.
The electrical resistivity (\(\rho\)) of the plate material relates to how easily current flows *through* the plate material itself. It does not directly influence the electric field or charge storage *between* the plates, which is the essence of capacitance.
Comparing C1 and C2
In this problem, both capacitors have the same plate area (\(A\)) and the same plate separation (\(d\)). Assuming the dielectric medium between the plates is the same (e.g., vacuum or air), the permittivity (\(\epsilon\)) is also the same for both.
Therefore, the capacitance of C1 is:
\(C_1 = \frac{\epsilon A}{d}\)
And the capacitance of C2 is:
\(C_2 = \frac{\epsilon A}{d}\)
Since \(\epsilon\), \(A\), and \(d\) are the same for both capacitors, we can conclude that:
\(C_1 = C_2\)
The electrical resistivity of the plate materials (\(\rho_g\) and \(\rho_a\)) does not affect the capacitance in this standard parallel-plate capacitor model.
Evaluating the Options
- Option 1: \(C_1 > C_2\). Incorrect, as we found \(C_1 = C_2\).
- Option 2: \(C_2 > C_1\). Incorrect, as we found \(C_1 = C_2\).
- Option 3: \(C_1 \rho_a = C_2 \rho_g\). Incorrect. Resistivity of the plates is not directly related to capacitance in this way.
- Option 4: \(C_1 = C_2\). Correct, as explained above.
Thus, the correct relation is \(C_1 = C_2\).
Revision Table: Parallel Plate Capacitor
| Parameter | Symbol | Effect on Capacitance (C) |
|---|---|---|
| Plate Area | A | C is directly proportional to A (\(C \propto A\)) |
| Plate Separation | d | C is inversely proportional to d (\(C \propto 1/d\)) |
| Permittivity of Dielectric | \(\epsilon\) | C is directly proportional to \(\epsilon\) (\(C \propto \epsilon\)) |
| Plate Material Resistivity | \(\rho\) | Does NOT affect C in the standard model |
Additional Information: Dielectric and Conductors
In a parallel-plate capacitor, the charge accumulates on the inner surfaces of the conducting plates. The electric field is primarily confined to the region between the plates, which is filled with a dielectric material (or vacuum/air). The dielectric material, although not a perfect conductor, can be polarized by the electric field, which increases the ability of the capacitor to store charge for a given voltage. This ability is quantified by its permittivity \(\epsilon\). The plate material's resistivity \(\rho\) determines how easily current flows *through* the plates or if there are significant voltage drops along the plates themselves, which is usually negligible in ideal capacitor theory unless dealing with very high frequencies or specific non-ideal scenarios. For standard DC or low-frequency AC circuits, the plates are considered equipotentials, and their resistivity does not influence the capacitance value \(\frac{\epsilon A}{d}\).
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