Q.In a parallel plate capacitor with air between the plates, each plate has an area of $6 \times 10^{-3}\ \text{m}^2$ and the distance between the plates is $3\ \text{mm}$. Calculate the capacitance of the capacitor. If this capacitor is connected to a $100\ \text{V}$ supply, what is the charge on each plate of the capacitor?

Concept: Parallel Plate Capacitance and Plate Separation — The capacitance of a parallel plate capacitor depends only on geometry and the dielectric between the plates.

Step 1: Capacitance formula

For a parallel plate capacitor with air (dielectric constant $\kappa = 1$),

where $\varepsilon_0 = 8.85 \times 10^{-12}\ \text{F/m}$, $A = 6 \times 10^{-3}\ \text{m}^2$, and $d = 3\ \text{mm} = 3 \times 10^{-3}\ \text{m}$.

Step 2: Calculate $C$

$C = \frac{(8.85 \times 10^{-12})(6 \times 10^{-3})}{3 \times 10^{-3}} = 1.77 \times 10^{-11}\ \text{F} = 17.7\ \text{pF}$ …

The core idea here is that a parallel plate capacitor stores charge in proportion to the voltage applied, and the constant of proportionality — the capacitance — is fixed by the physical construction: the area of the plates and how far apart they are. Air between the plates means we use the permittivity of free space, $\varepsilon_0$, with no dielectric factor.

Let's walk through it.

1. Recall the formula for a parallel plate capacitor. For two plates of area $A$, separated by distance $d$ in vacuum (or air, to excellent approximation), the capacitance is

where $\varepsilon_0 = 8.85 \times 10^{-12}\ \text{F/m}$ is the permittivity of free space.

This formula comes from Gauss's law: the electric field between the plates is uniform ($E = \sigma / \varepsilon_0$), the potential difference is $V = Ed$, and the charge is $Q = \sigma A$. Combining gives $C = Q/V = \varepsilon_0 A / d$.

2. Plug in the given numbers. Area: $A = 6 \times 10^{-3}\ \text{m}^2$ Separation: $d = 3\ \text{mm} = 3 \times 10^{-3}\ \text{m}$ So

Simplify step by step:

That is $17.7\ \text{pF}$ (picofarads). …

Method: Computing Capacitance and Charge for a Parallel Plate Capacitor from Geometry

This method finds the capacitance of an air-filled parallel plate capacitor from its physical dimensions, then the charge once a voltage is applied.

Steps

Step 1: Identify the geometric quantities and convert units

Note the plate area $A$ and separation $d$; convert every length to SI (metres) before substituting — mixing cm/mm with metres is the most common source of error here.

Step 2: Apply the parallel-plate formula

$C = \frac{\varepsilon_0 A}{d}$ …