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:

Capacitance depends only on geometric factors: directly proportional to area $A$, inversely proportional to plate separation $d$, and permittivity of medium. It is independent of total charge $Q$ or potential $V$.

Answer 1

Concept:
• A parallel plate capacitor consists of two conducting parallel plates, each of area $A$, separated by small distance $d$.

• Plates carry equal and opposite surface charge densities $\sigma = +\frac{Q}{A}$ and $-\sigma = -\frac{Q}{A}$.

• Electric field in outer regions is zero, while uniform electric field exists in region between plates.

Step 1: Electric Field Between Plates
Electric field due to a single infinite thin plane sheet of charge is $E = \frac{\sigma}{2\varepsilon_0}$.
In the region between positively charged plate ($+\sigma$) and negatively charged plate ($-\sigma$):
Both electric fields point in the same direction (from positive to negative plate).
Total electric field $E$:
\[ E = \frac{\sigma}{2\varepsilon_0} + \frac{\sigma}{2\varepsilon_0} = \frac{\sigma}{\varepsilon_0} \]
Substitute surface charge density $\sigma = \frac{Q}{A}$:
\[ E = \frac{Q}{\varepsilon_0 A} \]

Step 2: Potential Difference Between Plates
Since electric field $E$ is uniform over distance $d$, potential difference $V$ between plates is:
\[ V = E \cdot d = \left( \frac{Q}{\varepsilon_0 A} \right) d \]

Step 3: Capacitance Calculation
Capacitance $C$ is defined as ratio of total charge $Q$ to potential difference $V$:
\[ C = \frac{Q}{V} \]
Substitute expression for $V$:
\[ C = \frac{Q}{\frac{Q d}{\varepsilon_0 A}} = \frac{\varepsilon_0 A}{d} \]

Step 4: Conclusion
The capacitance of a parallel plate capacitor filled with air is $C = \frac{\varepsilon_0 A}{d}$.