PHYS 004B

Hour 17

Capacitance

Two expressions for capacitance:

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

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

Capacitors in Series

Rules for capacitors in series:

The charge on each capacitor is the same, even if each capacitor has a different capacitance.

$$ q_1= q_2 = q_3 $$

The equivalent capacitance is the reciprocal of the sum of the reciprocals of the individual capacitances.

$$ \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \dots + \frac{1}{C_n} $$

Capacitors in Parallel

The potential difference across each capacitor is the same for capacitors in parallel.

$$ \Delta{V_1} = \Delta{V_2} = \Delta{V_3} $$

The equivalent capacitance is the sum of the individual capacitances.

$$ C_{eq} = C_1 + C_2 + \dots + C_n $$

Hour 18

Canvas - Video 18a

Youtube - Video 18a

Electric Potential Energy

$$ U = \frac{1}{2}Q\Delta{V} $$

Equivalently,

$$ U = \frac{1}{2}C(\Delta{V}^2) $$

Or alternatively,

$$ U = \frac{1}{2}\frac{Q^2}{C} $$

Youtube - Video 18b

Canvas - Video 18b

A dielectric is an insulator that is placed between the plates of a capacitor. It increases the capacitance of the capacitor.

$$ C = \kappa * \frac{\varepsilon_0 A}{d} $$

Kappa is the dielectric constant, and is always greater than 1.

Capacitance with a dielectric:

$$ C = \kappa * C_0 $$

Where $C_0$ is the capacitance without a dielectric, and $\kappa$ is the dielectric constant.

Hour 19

Canvas - Video 19a

Example 1

Consider a scenario where we disconnect a capacitor from a battery, and push the two plates closer together.

	Initial	Final	Change
Spacing between the plates	D	d	↓
Potential difference between the plates ($\Delta{V}$)			↓
Charge ($Q$)	$Q_0$	$Q_0$	=
Surface Charge Density ($\sigma$)	$\sigma_0$	$\sigma_0$	=
Electric Field ($E$)			=
Capacitance ($C$)			↑

The potential difference between the plates decreased, and the capacitance increased.

Example 2

In the next example, we leave it connected to the battery the entire time.

	Initial	Final	Change
Spacing between the plates	D	d	↓
Potential difference between the plates ($\Delta{V}$)			=
Charge ($Q$)	$Q_0$	$Q_0$	↑
Surface Charge Density ($\sigma$)	$\sigma_0$	$\sigma_0$	↑
Electric Field ($E$)			↑
Capacitance ($C$)			↑

The capacitance increased because the distance decreased. The relevant formula is $C = \frac{\varepsilon_0 A}{d}$.

The charge had to go up because $Q = C\Delta{V}$, and $\Delta{V}$ is constant.

The area didn't change, but the charge did, so the surface charge density increased as well.