Physics 219 Question 1 January

Transcription

1 Lecture 6-16 Physics 219 Question 1 January A (non-ideal) battery of emf 1.5 V and internal resistance 5 Ω is connected to a light bulb of resistance 50 Ω. How much power is delivered to the light bulb? a) 0.75 W b) 0.19 W c) W d) W e) W

2 Lecture 6-26 Q +Q -Q ΔQ e - Q ΔQ Capacitors A capacitor is a device that is capable of storing electric charges and thus electric potential energy. => charging Its purpose is to release them later in a controlled way. => discharging Capacitors are used in vast majority of electrical and electronic devices. Typically made of two conductors and, when charged, each holds equal and opposite charges.

3 Lecture 6-36 Capacitance Capacitor plates hold charge Q The capacitance C of a capacitor is a measure of how much charge Q it can store for a given potential difference V between the plates. Ο V (volts) [ C] The two conductors hold charge +Q and Q, respectively. Expect [ Q] [ V] Potential difference = V = = electromotive force Let C Q V Q V Capacitance is an intrinsic property of the capacitor. = = Coulomb F Volt farad

4 Lecture 6-46 Steps to calculate capacitance C 1. Put charges Q and -Q on the two plates, respectively. 2. Calculate the electric field E between the plates due to the charges Q and -Q, e.g., by using Gauss s law. 3. Calculate the potential difference V between the plates due to the electric field E. 4. Calculate the capacitance of the capacitor by dividing the charge by the potential difference, i.e., C = Q/ V.

5 Lecture 6-56 Example: (Ideal) Parallel-Plate Capacitor 1. Put charges +q and q on plates of area A and separation d. 2. Calculate E by Gauss s Law S E ds = E A= From which we obtain: E = q A 3. Calculate V by 0 q enclosed 0 ( Eq.1) Area of Gaussian surfaces = A a b V = V V = Ed ( Eq.2) b a

6 Lecture To obtain C, divide q by V and substitute for V and E from Eq.1 and Eq. 2 q q q A 0 C = = = = V Ed d q 0 A d q is indeed prop. to V C is prop. to A C is inversely prop. to d

7 Lecture 6-76 Numerical magnitudes Let s say: Area A= 1 cm², separation d=1 mm C Then 0 A = d = ( C / Nm ) 10 ( m ) 3 10 ( ) m ( / ) ( ) = C N m = F = pf This is on the order of 1pF (pico farad).

8 Lecture 6-86 Prefixes for Powers of 10 Multiple Prefix Abbreviation 3 10 milli m 6 10 micro μ 9 10 nano n pico p μf nf pf = 10 = 10 = F F F Generally the values of typical capacitors are more conveniently measured in μf, nf or pf.

9 Lecture 6-96 Physics 219 Question 2 January 30, You wish to make a parallel plate capacitor with C = 8.85 μf. If the plates are to be square and the distance between the plates is to be 0.1 mm (=10-4 m), what would the edge length have to be? a) m b) 100 m c) 10 m d) 316 m e) 3.4 m 0 = C 2 /(Nm 2 ), C = d 0 A

10 Lecture Capacitors in Parallel V is common q q = = = q V C 1 C 2 C 3 Equivalent Capacitor holds the same charge as C 1, C 2, C 3 hold together: Thus: C q= q1+ q2 + q3 eq q = = V q + q + q V

11 Lecture C eq q + q + q = C1+ C2 V = + C 3 If d = d = d the above result is consistent with our earlier parallel-plate capacitor result of C A 0 = d A

12 Lecture Capacitors in Series q i q V CV CV.. CV s common = C 1 1= 2 2= = i i Equivalent Capacitor is the capacitor which holds the same unique charge which C 1 or C 2 or C 3 hold individually, that is q1 = q2 = q3 = q q q Ceq = = and V = V1+ V2 + V3 V V1+ V2 + V3 After inverting both side of the above equation, we obtain: 1 V + 1 V + 2 V = = + + C q C C C eq Consistent with parallelplate capacitor result of C = d d 1 or d C 0A 1

13 Lecture Summary of Capacitor Combination Series: = C C C C eq Each Capacitor C i holds the same charge: q i =q k =q=q eq Voltages add up to V=V 1 +V 2 +.V i Parallel: C = C1+ C2+ C eq Voltages are same on each C i V i =V k =V eq Charges add up. qi = qeq

14 Lecture Physics 219 Question 3 January 30, Below, all capacitors are identical. Which combination has the largest equivalent capacitance? a b c d e

15 Lecture Energy of a charged capacitor How much energy is stored in a charged capacitor? Calculate the work required (usually provided by a battery) to charge a capacitor to Q Calculate incremental work W i needed to move charge Δq from negative plate to the positive plate at voltage V i. V i Δq q Δ V i = qi C qi q i ( / ) W = V Δ q = q C Δq i i i i i V Total work U is then the area under the V-Q curve. qi 1 1 Q U = Wi = Δ qi = qdq= C C 2 C Q 0 2 V i = U 2 1 QV Q CV 2 2 2C 2 = = =

16 Lecture Where is the energy stored? Energy is stored in the electric field itself. Think of the energy needed to charge the capacitor as being the energy needed to create the electric field. U Q 1 Q 1 Q ( ) 2 = = = 0 Ad 2 C 2 ( A / d) 2 A 0 0 The electric field is given by: Q E = σ = A 1 U = E 2 0Ad The energy density u in the field is given by: This is the energy density, u, of the electric field. u U U = = volume Ad = 0E Units: 3 J / m