Notes
Conductors in electrostatic equilibrium
- The electric field inside a conductor is zero.
- Any excess charge sits on the surface.
- The field just outside is perpendicular to the surface: $\displaystyle E = \frac{\sigma}{\varepsilon_0}$.
- The whole conductor is at one potential. Charge piles up most at sharp points.
- A hollow conductor shields its interior from outside fields (a Faraday cage).
Capacitance
- $\displaystyle C = \frac{Q}{\Delta V}$ (farads). It depends only on geometry and the material between the conductors.
- Parallel plates: $\displaystyle C = \frac{\kappa\varepsilon_0A}{d}$.
- Other geometries: find $E$ with Gauss's law, find $\displaystyle \Delta V = \int E\,dr$, then $\displaystyle C = \frac{Q}{\Delta V}$.
- Energy stored: $\displaystyle U = \frac{1}{2}C(\Delta V)^2 = \frac{Q^2}{2C} = \frac{1}{2}Q\Delta V$. Energy density in a field: $\displaystyle u = \frac{1}{2}\varepsilon_0E^2$.
Dielectrics
A dielectric (constant $\kappa > 1$) between the plates reduces the field and increases capacitance by a factor $\kappa$.
- Battery still connected: $\Delta V$ is fixed, so $Q$ increases.
- Battery disconnected: $Q$ is fixed, so $\Delta V$ decreases.
Combinations
Parallel: $C_{eq} = C_1 + C_2$ (same $\Delta V$). Series: $\displaystyle \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2}$ (same $Q$).
Practice
Otherwise free charges would keep moving.
$\displaystyle C \propto \frac{1}{d}$.
$Q = C\Delta V$.
$\displaystyle \frac{1}{2}(10 \times 10^{-6})(144)$.
$Q$ is fixed and $C$ increases, so $\displaystyle \Delta V = \frac{Q}{C}$ falls.
$\displaystyle \frac{1}{C} = \frac{1}{3} + \frac{1}{6} = \frac{1}{2}$.
Sharp points have the smallest radius of curvature.
$\Delta V$ is fixed and $C$ decreases, so $Q = C\Delta V$ decreases.
Free response
Write your answer first, then open the scoring guide.
Free-response question
A spherical capacitor has an inner conducting sphere of radius $a$ with charge $+Q$ and an outer conducting shell of radius $b$ with charge $-Q$.
(a) Use Gauss's law to find $E$ between the spheres ($a < r < b$).
Show the answer and scoring guide
A spherical Gaussian surface encloses $+Q$: $\displaystyle E(4\pi r^2) = \frac{Q}{\varepsilon_0} \Rightarrow E = \frac{Q}{4\pi\varepsilon_0r^2}$.
Check off the points you earned:
(b) Find the potential difference between the spheres.
Show the answer and scoring guide
$\displaystyle \Delta V = \int_a^b \frac{Q}{4\pi\varepsilon_0r^2}dr = \frac{Q}{4\pi\varepsilon_0}\left(\frac{1}{a} - \frac{1}{b}\right)$.
Check off the points you earned:
(c) Find the capacitance.
Show the answer and scoring guide
$\displaystyle C = \frac{Q}{\Delta V} = \frac{4\pi\varepsilon_0ab}{b - a}$.
Check off the points you earned: