Question

Derive an expression for the capacitance of a parallel plate capacitor of plate area \(A\) and plate separation \(d\) with air present between the plates.

Hint:

For a parallel plate capacitor: \[ C=\frac{\varepsilon_0A}{d} \] Remember: \[ C \propto A \] and \[ C \propto \frac{1}{d} \] Thus, increasing plate area increases capacitance, whereas increasing plate separation decreases capacitance.

Answer 1

Concept: A capacitor is a device used for storing electric charge and electrical energy. A parallel plate capacitor consists of two large conducting plates placed parallel to each other and separated by a small distance. When the plates are connected to a source of potential difference, equal and opposite charges develop on the two plates. The capacitance of a capacitor is defined as the ratio of the charge stored on either plate to the potential difference between the plates. \[ C=\frac{Q}{V} \] where
• \(C\) = capacitance,
• \(Q\) = charge on either plate,
• \(V\) = potential difference between the plates.

Step 1: Consider a parallel plate capacitor. Let
• Area of each plate \(=A\),
• Separation between the plates \(=d\),
• Charge on the plates \(=\pm Q\). The surface charge density on the plates is \[ \sigma=\frac{Q}{A} \]

Step 2: Determine the electric field between the plates. The electric field due to a single charged conducting plate is \[ E=\frac{\sigma}{2\varepsilon_0} \] where \(\varepsilon_0\) is the permittivity of free space. Since the capacitor consists of two oppositely charged plates, the electric fields between the plates add up. Therefore, \[ E=\frac{\sigma}{2\varepsilon_0}+\frac{\sigma}{2\varepsilon_0} \] \[ E=\frac{\sigma}{\varepsilon_0} \] Substituting \[ \sigma=\frac{Q}{A} \] we obtain \[ E=\frac{Q}{\varepsilon_0 A} \]

Step 3: Calculate the potential difference between the plates. The potential difference between two points separated by distance \(d\) in a uniform electric field is \[ V=Ed \] Substituting the value of \(E\), \[ V=\frac{Q}{\varepsilon_0 A}\,d \] \[ V=\frac{Qd}{\varepsilon_0 A} \]

Step 4: Determine the capacitance. Using the definition \[ C=\frac{Q}{V} \] Substituting the value of \(V\), \[ C=\frac{Q}{\dfrac{Qd}{\varepsilon_0 A}} \] \[ C=\frac{\varepsilon_0 A}{d} \] Final Result: Hence, the capacitance of a parallel plate capacitor having air between the plates is \[ \boxed{C=\frac{\varepsilon_0 A}{d}} \] Observations from the formula:
• Capacitance is directly proportional to the area of the plates.
• Capacitance is inversely proportional to the separation between the plates.
• Capacitance depends upon the nature of the medium between the plates. A larger plate area increases charge storage capacity, while increasing the separation decreases capacitance.