AP Physics 2 / Unit 10: Electric Force, Field, and Potential
10.6 Capacitors
Notes
$$C=\frac{Q}{V},\qquad C=\frac{\kappa\varepsilon_0A}{d},\qquad U=\frac{1}{2}CV^2$$
- Larger plates or a smaller gap increase capacitance; dielectrics increase it by $\kappa$.
Practice
Halving the plate separation makes $C$:
$\displaystyle C\propto\frac{1}{d}$.
$C=2$ μF at 10 V stores charge:
$Q=CV$.
Energy stored in that capacitor:
$\displaystyle \frac{1}{2}(2\times10^{-6})(100)$.
Ready for more? Take the Unit 10 test.