Physics 2113 Lecture: 16 MON 23 FEB Jonathan Dowling Physics 2113 Lecture: 16 MON 23 FEB Capacitance I

Capacitors and Capacitance Capacitor: any two conductors, one with charge +Q, other with charge –Q –Q Potential DIFFERENCE between conductors = V +Q Uses: storing and releasing electric charge/energy. Most electronic capacitors: micro-Farads (μF), pico-Farads (pF) — 10–12 F New technology: compact 1 F capacitors Q = CV where C = capacitance For fixed Q the larger the capacitance the larger the stored voltage. Units of capacitance: Farad (F) = Coulomb/Volt

Capacitance +Q –Q (We first focus on capacitors Capacitance depends only on GEOMETRICAL factors and on the MATERIAL that separates the two conductors e.g. Area of conductors, separation, whether the space in between is filled with air, plastic, etc. +Q –Q (We first focus on capacitors where gap is filled by AIR!)

Capacitance depends only on GEOMETRICAL factors and on the MATERIAL that separates the two conductors. It does not depend on the voltage across it or the charge on it at all. Same (b) Same

Capacitors Electrolytic (1940-70) Paper (1940-70) Electrolytic (new) Variable air, mica Ceramic (1930 on) Tantalum (1980 on) Mica (1930-50)

Parallel Plate Capacitor We want capacitance: C = Q/V  E field between conducting plates: Area of each plate = A Separation = d charge/area = σ = Q/A Relate E to potential difference V: +Q –Q σ d What is the capacitance C ?

Capacitance and Your iPhone!

Parallel Plate Capacitor — Example A huge parallel plate capacitor consists of two square metal plates of side 50 cm, separated by an air gap of 1 mm What is the capacitance? C = ε0A/d = (8.85 x 10–12 F/m)(0.25 m2)/(0.001 m) = 2.21 x 10–9 F (Very Small!!) Lesson: difficult to get large values of capacitance without special tricks!

Isolated Parallel Plate Capacitor: ICPP A parallel plate capacitor of capacitance C is charged using a battery. Charge = Q, potential voltage difference = V. Battery is then disconnected. If the plate separation is INCREASED, does the capacitance C: (a) Increase? (b) Remain the same? (c) Decrease? +Q –Q Q is fixed! d increases! C decreases (= ε0A/d) V=Q/C; V increases.

Parallel Plate Capacitor & Battery: ICPP A parallel plate capacitor of capacitance C is charged using a battery. Charge = Q, potential difference = V. Plate separation is INCREASED while battery remains connected. +Q –Q V is fixed constant by battery! C decreases (=ε0A/d) Q=CV; Q decreases E = σ/ε0 = Q/ε0A decreases Does the Electric Field Inside: (a) Increase? (b) Remain the Same? (c) Decrease? Battery does work on capacitor to maintain constant V!

Spherical Capacitor Concentric spherical shells: What is the electric field inside the capacitor? (Gauss’ Law) Radius of outer plate = b Radius of inner plate = a Concentric spherical shells: Charge +Q on inner shell, –Q on outer shell Relate E to potential difference between the plates:

Spherical Capacitor What is the capacitance? C = Q/V = Radius of outer plate = b Radius of inner plate = a Concentric spherical shells: Charge +Q on inner shell, –Q on outer shell Isolated sphere: let b >> a,

Cylindrical Capacitor What is the electric field in between the plates? Gauss’ Law! Radius of outer plate = b Radius of inner plate = a Length of capacitor = L +Q on inner rod, –Q on outer shell Relate E to potential difference between the plates: cylindrical Gaussian surface of radius r

Summary Any two charged conductors form a capacitor. Capacitance : C= Q/V Simple Capacitors: Parallel plates: C = ε0 A/d Spherical: C = 4πε0 ab/(b-a) Cylindrical: C = 2πε0 L/ln(b/a)]

Parallel plates: C = ε0 A/d Spherical: Cylindrical: C = 2πε0 L/ln(b/a)] Hooked to battery V is constant. Q=VC (a) d increases -> C decreases -> Q decreases (b) a inc -> separation d=b-a dec. -> C inc. -> Q increases (c) b increases -> separation d=b-a inc.-> C dec. -> Q decreases

Capacitors in Parallel: V=Constant An ISOLATED wire is an equipotential surface: V = Constant Capacitors in parallel have SAME potential difference but NOT ALWAYS same charge! VAB = VCD = V Qtotal = Q1 + Q2 CeqV = C1V + C2V Ceq = C1 + C2 Equivalent parallel capacitance = sum of capacitances V = VAB = VA –VB V = VCD = VC –VD VA VB VC VD A B C D C1 C2 Q1 Q2 PAR-V (Parallel V the Same) ΔV=V Ceq Qtotal PARALLEL: V is same for all capacitors Total charge = sum of Q

Capacitors in Series: Q=Constant Isolated Wire: Q=Q1=Q2=Constant Q1 = Q2 = Q = Constant VAC = VAB + VBC A B C C1 C2 Q1 Q2 SERI-Q (Series Q the Same) Q = Q1 = Q2 SERIES: Q is same for all capacitors Total potential difference = sum of V Ceq

PARALLEL: V is same for all capacitors Total charge = sum of Q SERIES: Q is same for all capacitors Total potential difference = sum of V Parallel: Voltage is same V on each but charge is q/2 on each since q/2+q/2=q. (b) Series: Charge is same q on each but voltage is V/2 on each since V/2+V/2=V.

“Compiling” in parallel and in series Cpar = C1 + C2 Vpar = V1 = V2 Qpar = Q1 + Q2 C1 C2 Q1 Q2 Ceq Qeq In series : 1/Cser = 1/C1 + 1/C2 Vser = V1  + V2 Qser= Q1 = Q2 C1 C2 Q1 Q2

Example: Parallel or Series? Parallel: Circuit Splits Cleanly in Two (Constant V) What is the charge on each capacitor? C1=10 μF C3=30 μF C2=20 μF 120V Qi = CiV V = 120V = Constant Q1 = (10 μF)(120V) = 1200 μC Q2 = (20 μF)(120V) = 2400 μC Q3 = (30 μF)(120V) = 3600 μC Note that: Total charge (7200 μC) is shared between the 3 capacitors in the ratio C1:C2:C3 — i.e. 1:2:3

Example: Parallel or Series Series: Isolated Islands (Constant Q) What is the potential difference across each capacitor? C1=10μF C2=20μF C3=30μF Q = CserV Q is same for all capacitors Combined Cser is given by: 120V Ceq = 5.46 μF (solve above equation) Q = CeqV = (5.46 μF)(120V) = 655 μC V1= Q/C1 = (655 μC)/(10 μF) = 65.5 V V2= Q/C2 = (655 μC)/(20 μF) = 32.75 V V3= Q/C3 = (655 μC)/(30 μF) = 21.8 V Note: 120V is shared in the ratio of INVERSE capacitances i.e. (1):(1/2):(1/3) (largest C gets smallest V)

Example: Series or Parallel? Neither: Circuit Compilation Needed! 10μF 5μF 10V In the circuit shown, what is the charge on the 10μF capacitor? 10 μF 10μF 10V The two 5μF capacitors are in parallel Replace by 10μF Then, we have two 10μF capacitors in series So, there is 5V across the 10 μF capacitor of interest Hence, Q = (10μF )(5V) = 50μC