Parallel plate capacitance depends on area, separation, and dielectric

Capacitance & Charge/Discharge — AQA A-Level Physics

C = \frac{A \varepsilon_0 \varepsilon_r}{d}

$C$: capacitanceThe chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C). stored per unit potential difference across a capacitor. Measured in farads (F). (F)
$A$: cross-sectional area of one plate ($m^{2}$)
$varepsilon_0$: permittivity of free space (8.85 x 10^-12 F m^-1)
$varepsilon_r$: relative permittivity of the dielectricAn insulating material placed between the plates of a capacitor that increases its capacitance by reducing the electric field strength for a given chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C).. (dimensionless)
$d$: separation between the plates (m)

Larger plate area A: more chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C). can be stored, higher capacitanceThe charge stored per unit potential difference across a capacitor. Measured in farads (F)..
Smaller plate separation d: stronger field for the same p.d., higher capacitanceThe charge stored per unit potential difference across a capacitor. Measured in farads (F)..
Higher relative permittivity: dielectricAn insulating material placed between the plates of a capacitor that increases its capacitance by reducing the electric field strength for a given charge. reduces the effective field, higher capacitance.
For square plates of length L, the area is $L^{2}$.

Worked Example

Capacitor 1 has square plates of length L, separation d, and is filled with a dielectric. Capacitor 2 has square plates of length 3L, separation 3d, and air as its dielectric. Both have the same capacitance. Find the relative permittivity of the dielectric in capacitor 1.

Show Solution

Write capacitance for each

$$C_1 = \frac{L^2 \varepsilon_0 \varepsilon_r}{d} \qquad C_2 = \frac{(3L)^2 \varepsilon_0}{3d} = \frac{9L^2 \varepsilon_0}{3d}$$

Equate and simplify

$$\frac{L^2 \varepsilon_0 \varepsilon_r}{d} = \frac{9L^2 \varepsilon_0}{3d}$$

Cancel $L^2$, $\varepsilon_0$, and $d$:

$$\varepsilon_r = \frac{9}{3} = 3$$

Answer

$\varepsilon_r = 3$

Examiner Tips and Tricks

A is the area of one plate, not both.
Do not multiply by 2.