AP Physics 2 Exam Question:

A parallel plate capacitor consists of two conducting plates of area $A$, separated by a distance $d$. The plates are connected to a battery with voltage $V$. The dielectric material with dielectric constant $κ$ is inserted between the plates.

1. Derive the formula for the capacitance $C$ of the parallel plate capacitor.

Answer:

To derive the formula for the capacitance $C$ of a parallel plate capacitor, we can start with the definition of capacitance, which relates the charge stored on the plates ($Q$) to the potential difference across the plates ($V$):

$C = \frac{Q}{V}$

To find the charge $Q$, we can consider the electric field between the plates. The electric field strength ($E$) between the plates of a parallel plate capacitor is given by:

$E = \frac{V}{d}$

The electric field is directly proportional to the charge density ($σ$) on the plates, which is defined as the charge per unit area:

$σ = \frac{Q}{A}$

By combining these equations, we can relate the charge $Q$ to the area $A$ and the distance $d$:

$Q = Aσ$

Substituting this expression for $Q$ into the formula for capacitance, we get:

$C = \frac{Aσ}{V} = \frac{A\cdot \frac{Q}{A}}{V} = \frac{Q}{V}$

Since $Q = CV$, we can rewrite the formula for capacitance as:

$C = \frac{Q}{V}$

So, the formula for the capacitance $C$ of a parallel plate capacitor is:

$C = \frac{ε_0A}{d}$

where $ε_0$ is the permittivity of the vacuum (or free space), approximately $8.85 \times 10^{-12}$ F/m.

Note: Inserting a dielectric material between the plates increases the capacitance by a factor of the dielectric constant $κ$, so the formula for capacitance with a dielectric material becomes:

$C = κ\cdot\frac{ε_0A}{d}$