Capacitors Physics 1161: Lecture 05 Textbook Sections 20-5 – 20-6

Capacitance Practice How much charge is on a 0.9 F capacitor which has a potential difference of 200 Volts? Q = CV How much energy is stored in this capacitor? U = ½ Q V = (0.9)(200) = 180 Coulombs = ½ (180) (200) = 18,000 Joules!

Parallel Plate Capacitor C =  0 A/d Calculate the capacitance of a parallel plate capacitor made from two large square metal sheets 1.3 m on a side, separated by 0.1 m. A d A

A parallel plate capacitor is given a charge q. The plates are then pulled a small distance further apart. What happens to the charge q on each plate of the capacitor? 1.Increases 2.Constant 3.Decreases +q-q d pull

A parallel plate capacitor is given a charge q. The plates are then pulled a small distance further apart. What happens to the charge q on each plate of the capacitor? 1.Increases 2.Constant 3.Decreases +q-q d pull Remember charge is real/physical. There is no place for the charges to go.

A parallel plate capacitor is given a charge q. The plates are then pulled a small distance further apart. Which of the following apply to the situation after the plates have been moved? 1)The capacitance increases True False 2)The electric field increases True False 3)The voltage between the plates increases True False 4)The energy stored in the capacitor increases True False Checkpoint +q-q d pull C =  0 A/dC decreases! E= Q/(  0 A)Constant V= Ed U = ½ QV 40% 50% 55% 70%

Two identical parallel plate capacitors are shown in end-view in A) of the figure. Each has a capacitance of C. A) B) If the two are joined as in (B) of the figure, forming a single capacitor, what is the final capacitance? 1) 2C 2) C 3) C/2 Checkpoint

Physics 1161: Lecture 4, Slide 8 Parallel Practice A 4  F capacitor and 6  F capacitor are connected in parallel and charged to 5 volts. Calculate C eq, and the charge on each capacitor. C4 C6 Ceq 5 V 0 V 5 V 0 V 5 V 0 V Ceq = C4+C6 Q4 = C4 V4 Q6 = C6 V6 Qeq = Ceq Veq = 4  F+6  F = 10  F = (4  F)(5 V) = 20  C = (6  F)(5 V) = 30  C = (10  F)(5 V) = 50  C = Q4+ Q6 V = 5 V

Series Practice Ceq C4 C Q -Q +Q -Q +Q -Q A 4  F capacitor and 6  F capacitor are connected in series and charged to 5 volts. Calculate C eq, and the charge on the 4  F capacitor. 5 V 0 V 5 V 0 V Q = CV

Comparison: Series vs. Parallel Series Can follow a wire from one element to the other with no branches in between. Parallel Can find a loop of wire containing both elements but no others (may have branches). C1 C2 C1

A circuit consists of three initially uncharged capacitors C1, C2, and C3, which are then connected to a battery of emf . The capacitors obtain charges q1, q2,q3, and have voltages across their plates V1, V2, and V3. C 2 C 3 C 1 E + - -q2 +q1 -q1 +q3 -q3 +q2 V1 V2 V3 1)q1 = q2 2)q2 = q3 3)V2 = V3 4)E = V1 5)V1 < V2 6)Ceq > C1 Checkpoint

A circuit consists of three initially uncharged capacitors C1, C2, and C3, which are then connected to a battery of emf . The capacitors obtain charges q1, q2,q3, and have voltages across their plates V1, V2, and V3. C 2 C 3 C 1 E + - -q2 +q1 -q1 +q3 -q3 +q2 V1 V2 V3 1) q1 = q2 Not necessarily C1 and C2 are NOT in series. 2) q2 = q3 Yes! C2 and C3 are in series. Checkpoint 55% 35%

C 2 C 3 C 1 E + - -q2 +q1 -q1 +q3 -q3 +q2 V1 V2 V3 3) V2 = V3Not necessarily, only if C2 = C3 4) E = V1 10V 0V 7V?? Yes! Both ends are connected by wires A circuit consists of three initially uncharged capacitors C1, C2, and C3, which are then connected to a battery of emf . The capacitors obtain charges q1, q2,q3, and have voltages across their plates V1, V2, and V3. Checkpoint 55% 45%

C 2 C 3 C 1 E + - -q2 +q1 -q1 +q3 -q3 +q2 V1 V2 V3 5) V1 < V2Nope, V1 > V2. (E.g. V1 = 10-0, V2 =10-7 6) Ceq > C1 10V 0V 7V?? Yes! C1 is in parallel with C23 (Ceq = C1 + C23) A circuit consists of three initially uncharged capacitors C1, C2, and C3, which are then connected to a battery of emf . The capacitors obtain charges q1, q2,q3, and have voltages across their plates V1, V2, and V3. Checkpoint 15% 50%

Battery Batteries produce electrons through chemical reactions. htm htm

Capacitance C = Q/V Parallel Plate: C =  0 A/d Dielectric: C = kC o (k is dielectric constant) Capacitors in parallel: C eq = C 1 +C 2 Capacitors in series: C eq = 1/(1/C 1 +1/C 2 ) Batteries provide fixed potential difference Recap of Today’s Lecture