Title: InkPhysics / Capacitance
On every note of this deck, unless a note says otherwise:
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Front: CHECK YOURSELF
A 12 V supply charges a capacitor through a 10 kΩ resistor. State the current the moment the switch closes, and the current and capacitor pd after a long time, explaining each value.
Show a hint
At the first instant the capacitor opposes nothing; after a long time it opposes everything.
Back: At t = 0 the uncharged capacitor has no pd, so the whole 12 V sits across the resistor: I_0 = V/R = 12 / 10 000 = 1.2 mA.
After a long time the capacitor pd has climbed to 12 V, matching the supply.
With no pd left across the resistor the current is zero: the capacitor is fully charged and the circuit rests.

Front: DEFINITION
Define: Capacitance.
Back: The charge stored per unit potential difference: C = Q/V, in farads.

Front: CHECK YOURSELF
A 470 μF capacitor discharges through a 10 kΩ resistor. Find the time constant, the half-life, and the fraction of the charge remaining after 9.4 s.
Show a hint
Base units first; then notice what 9.4 s is in time constants.
Back: RC = 10 000 × 470 × 10^−6 = 4.7 s.
T_½ = 0.69RC = 0.69 × 4.7 = 3.2 s.
9.4 s is two time constants, so the fraction left is e^−2 = 0.135, about 13.5%: two rounds of keeping 37%.

Front: CHECK YOURSELF
A capacitor has plates of area 0.020 m² separated by 1.0 mm, with a dielectric of relative permittivity 2.5. Find its capacitance, and the charge and energy it stores at 12 V.
Show a hint
Geometry first, then Q = CV, then the energy form that uses what you now hold.
Back: C = Aε_0ε_r/d = (0.020 × 8.85 × 10^−12 × 2.5) / 1.0 × 10^−3 = 4.4 × 10^−10 F, about 0.44 nF.
Q = CV = 4.4 × 10^−10 × 12 = 5.3 × 10^−9 C.
E = 12CV^2 = 0.5 × 4.4 × 10^−10 × 144 = 3.2 × 10^−8 J: tiny numbers, so the farad rarely appears without a prefix.

Front: DEFINITION
Define: Time constant.
Back: The time RC for the charge on a discharging capacitor to fall to 1/e, about 37%, of its initial value.