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Capacitor Parallel Plates Charged plates each store + - charge. • Positive charge at higher potential • Negative charge at lower potential The charge storing device is called a capacitor. • Old term condenser + - Capacitance The measure of a device to store charge at a given voltage is its capacitance. Q C V The unit of capacitance is the farad (F). • 1F=1C/V • Farads are a large amount of capacitance • Microfarads (mF) and picofarads (pF) are common Isolated Charge A metal sphere with a 0.25 m radius is in a vacuum. Find the capacitance. Note there is only one surface, but the formula still works. The potential of the sphere is V kQ Q r 40 r The capacitance is C Q r 40 r V k For the sphere • C= (0.25 m) / (8.99 x 109 Nm2/C2) • C = 28 x 10-12 F = 28 pF. Parallel Plate Capacitor + A capacitor is made from two - plates. • Area A • Distance d The field is due to the charge on the plates • E = Q/A + The voltage is V = Ed. The capacitance is C Q Q Q V Ed Qd A C A d - Keyboarding A key on a keyboard is a parallel plate capacitor. A springy insulator separates the plates. A circuit measures the capacitance from each key. • If pressed d decreases • The capacitance increases • The circuit records the key Two Capacitors Q C1V + C2V Two capacitor charged by the C1 + - same potential are in parallel. Q1 C1V The voltage is the same across each capacitor. The charge will be equal to the total from both. C2 Q2 C2V The total capacitance is the sum of the individual capacitances. + - C Q C1V + C2V V V C C1 + C2 Series Capacitors Two capacitors connected is Q Q1 Q2 Q1 C1V + Q2 C2V - sequence with the ends separated by a potential are in series. The total charge on the middle section must be neutral. • Same charge on capacitors • Voltages sum to the total C1 + C2 - C Q Q 1 V1 + V2 Q / C1 + Q / C2 1 / C1 + 1 / C2 1 1 1 + C C1 C2 More and Less Two capacitors 1mF and 3 mF are placed in parallel and in series. Find the combined capacitance in both situations. In parallel the capacitances add directly. In series the inverses are summed, then inverted. 1 1 1 + C C1 C2 1 (μF)-1 + 1/3 (μF)-1 = 4/3 (μF)-1 C = 3/4 μF C C1 + C2 C = 1 μF + 3 μF = 4 μF next