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
Electric Potential Energy
$$ U = \frac{1}{2}Q\Delta{V} $$
Equivalently,
$$ U = \frac{1}{2}C(\Delta{V}^2) $$
Or alternatively,
$$ U = \frac{1}{2}\frac{Q^2}{C} $$
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
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.
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