1. zero. Where an electric field line crosses an equipotential surface, the angle between the field line and the equipotential is


 Warren Lucas
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5 Where an electric field line crosses an equipotential surface, the angle between the field line and the equipotential is 1. zero 2. between zero and not enough information given to decide
6 Where an electric field line crosses an equipotential surface, the angle between the field line and the equipotential is 1. zero 2. between zero and not enough information given to decide Since there is no change in V along the equipotential surface ΔV = E dr = 0 But this means that the vectors E and dr are perpendicular to each other.
7 What is the electric potential created by these three batteries connected as shown? 1. 1 Volt 2. 2 Volt 3. 3 Volt 4. 4 Volt 5. 5 Volt 6. 6 Volt 7. 7 Volt 8. 8 Volt
8 What is the electric potential created by these three batteries connected as shown? 1. 1 Volt 2. 2 Volt 3. 3 Volt 4. 4 Volt 5. 5 Volt 6. 6 Volt 7. 7 Volt 8. 8 Volt
9 Which potentialenergy graph describes this electric field?
10 Which potentialenergy graph describes this electric field?
11 1. is the same as the direction of the electric field at that point The direction of the electric potential gradient at a certain point is opposite to the direction of the electric field at that point 3. is perpendicular to the direction of the electric field at that point 4. not enough information given to decide
12 1. is the same as the direction of the electric field at that point The direction of the electric potential gradient at a certain point is opposite to the direction of the electric field at that point 3. is perpendicular to the direction of the electric field at that point 4. not enough information given to decide E = V
13 Understand first a simpler analog: topographic maps have contour lines that are the locus of points such that ground elevation is constant: h( r) = constant
14 When you follow a path on a topographic map that crosses contour lines, you are either climbing or descending. A path running parallel to contour lines is flat. DENSE contour lines mean STEEP terrain and SPARSE ones mean FLAT terrain. The gradient is in the direction of steepest ascent and its size is the rate of ascent.
15 Which set of equipotential surfaces matches this electric field?
16 Which set of equipotential surfaces matches this electric field?
17 Example: Determine the electric field of a dipole, far away from the dipole Use the result from last week: V = kp cos θ r 2 = kp x r 3 Ans: (COB) E x = kp r2 3x 2 E y = kp 3xy r 5 r 5 = kp 1 3 cos2 θ r 3 θ cos θ = 3kpsin r 3
18 Three charged, metal spheres of different radii are connected by a thin metal wire. The potential and electric field at the surface of each sphere are V and E. Which of the following is true? 1. V 1 = V 2 = V 3 and E 1 = E 2 = E 3 2. V 1 = V 2 = V 3 and E 1 > E 2 > E 3 3. V 1 > V 2 > V 3 and E 1 = E 2 = E 3 4. V 1 > V 2 > V 3 and E 1 > E 2 > E 3 5. V 3 > V 2 > V 1 and E 1 = E 2 = E 3
19 Three charged, metal spheres of different radii are connected by a thin metal wire. The potential and electric field at the surface of each sphere are V and E. Which of the following is true? 1. V 1 = V 2 = V 3 and E 1 = E 2 = E 3 2. V 1 = V 2 = V 3 and E 1 > E 2 > E 3 3. V 1 > V 2 > V 3 and E 1 = E 2 = E 3 4. V 1 > V 2 > V 3 and E 1 > E 2 > E 3 5. V 3 > V 2 > V 1 and E 1 = E 2 = E 3
20 Capacitors (slide here really to remind me of demo 1st) A system consisting of two conductors ( plates ) Now, equal magnitude, opposite sign charge, ±Q, produces field between them (a) Hence, potential difference V The more charge Q, the larger V This is a linear relation (V Q) V = 1 C Q (b)
21 The two conductors a and b are insulated from each other, forming a capacitor. You increase the charge on a to +2Q and increase the charge on b to 2Q, while keeping the conductors in the same positions. What effect does this have on the capacitance C? 1. C is multiplied by a factor of 4 2. C is multiplied by a factor of 2 3. C is unchanged 4. C is multiplied by a factor of 1/2 5. C is multiplied by a factor of 1/4
22 The two conductors a and b are insulated from each other, forming a capacitor. You increase the charge on a to +2Q and increase the charge on b to 2Q, while keeping the conductors in the same positions. What effect does this have on the capacitance C? 1. C is multiplied by a factor of 4 2. C is multiplied by a factor of 2 3. C is unchanged 4. C is multiplied by a factor of 1/2 5. C is multiplied by a factor of 1/4
23 You reposition the two plates of a capacitor so that the capacitance doubles. If the charges +Q and Q on the two plates are kept constant in this process, what happens to the potential difference V ab between the two plates? 1. V ab is multiplied by a factor of 4 2. V ab is multiplied by a factor of 2 3. V ab is unchanged 4. V ab is multiplied by a factor of 1/2 5. V ab is multiplied by a factor of 1/4
24 You reposition the two plates of a capacitor so that the capacitance doubles. If the charges +Q and Q on the two plates are kept constant in this process, what happens to the potential difference V ab between the two plates? 1. V ab is multiplied by a factor of 4 2. V ab is multiplied by a factor of 2 3. V ab is unchanged 4. V ab is multiplied by a factor of 1/2 5. V ab is multiplied by a factor of 1/4
25 Example: our demo d is approx 0.5mm, and A is approx 1.0 m 2 (COB) C F = 18 nf so how much charge did we put in? for 1 kv: Q = CV C (This ignores the dielectric constant of the mica)
26 Capacitor C 1 is connected across a battery of 5 V. An identical capacitor C 2 is connected across a battery of 10 V. Which one has the most charge? 1) C 1 2) C 2 3) both have the same charge 4) it depends on other factors +Q Q
27 Capacitor C 1 is connected across a battery of 5 V. An identical capacitor C 2 is connected across a battery of 10 V. Which one has the most charge? 1) C 1 2) C 2 3) both have the same charge 4) it depends on other factors +Q Q Since Q = C V and the two capacitors are identical, the one that is connected to the greater voltage has the most charge, which is C 2 in this case.
28 What must be done to a capacitor in order to increase the amount of charge it can hold (for a constant voltage)? 1) increase the area of the plates 2) decrease separation between the plates 3) decrease the area of the plates 4) either (1) or (2) 5) either (2) or (3) +Q Q
29 What must be done to a capacitor in order to increase the amount of charge it can hold (for a constant voltage)? 1) increase the area of the plates 2) decrease separation between the plates 3) decrease the area of the plates 4) either (1) or (2) 5) either (2) or (3) Since Q = C V, in order to increase the charge +Q Q that a capacitor can hold at constant voltage, one has to increase its capacitance. Since the capacitance is given by, that can be done by either increasing A or decreasing d.
30 A parallelplate capacitor initially has a voltage of 400 V and stays connected to the battery. If the plate spacing is now doubled, what happens? 1) the voltage decreases 2) the voltage increases 3) the charge decreases 4) the charge increases 5) both voltage and charge change +Q Q
31 A parallelplate capacitor initially has a voltage of 400 V and stays connected to the battery. If the plate spacing is now doubled, what happens? 1) the voltage decreases 2) the voltage increases 3) the charge decreases 4) the charge increases 5) both voltage and charge change Since the battery stays connected, the voltage must remain constant! Since +Q Q when the spacing d is doubled, the capacitance C is halved. And since Q = C V, that means the charge must decrease. Followup: How do you increase the charge?
32 A parallelplate capacitor initially has a potential difference of 400 V and is then disconnected from the charging battery. If the plate spacing is now doubled (without changing Q), what is the new value of the voltage? 1) 100 V 2) 200 V 3) 400 V 4) 800 V 5) 1600 V +Q Q
33 A parallelplate capacitor initially has a potential difference of 400 V and is then disconnected from the charging battery. If the plate spacing is now doubled (without changing Q), what is the new value of the voltage? 1) 100 V 2) 200 V 3) 400 V 4) 800 V 5) 1600 V Once the battery is disconnected, Q has to remain constant, since no charge can flow either to or from the battery. Since when the spacing d is doubled, the capacitance C is halved. And since Q = C V, that means the voltage must double. +Q Q
34 The textbook calculates the capacitance of concentric spheres. Do another example: L coaxial cylindrical conductors Use previous chapter result (also from homework): V (a) V (b) = λ 2πɛ 0 ln ( b a ) b a (COB) C = 2πɛ 0L ln(b/a) Again, notice:  only depends on geometry,  increases with size of plates,  increases as plates get close ( a b )
35 Connecting circuit components in Parallel and in Series Some circuit component Two circuit elements connected in series: the charge of one necessarily flows into the other: Two circuit elements connected in parallel: the potential differences across them are necessarily equal:
36 Equivalent capacitance of capacitors in parallel C 1 C 2 = C 1 (COB) C = C 1 + C 2
37 Equivalent capacitance of capacitors in series C 1 C 2 = C 1 (COB) 1 C = 1 C C 2 +Q Q E +Q Q E
38 Two capacitors are connected together as shown. What is the equivalent capacitance of the two capacitors as a unit? 1. C eq = 18 µf 2. C eq = 9 µf 3. C eq = 6 µf 4. C eq = 4 µf 5. C eq = 2 µf
39 Two capacitors are connected together as shown. What is the equivalent capacitance of the two capacitors as a unit? 1. C eq = 18 µf 2. C eq = 9 µf 3. C eq = 6 µf 4. C eq = 4 µf 5. C eq = 2 µf
40 Two capacitors are connected together as shown. If the charge on the 12 µf capacitor is 24 microcoulombs (24 µc), what is the charge on the 6 µf capacitor? µc µc µc µc 5. 6 µc
41 Two capacitors are connected together as shown. If the charge on the 12 µf capacitor is 24 microcoulombs (24 µc), what is the charge on the 6 µf capacitor? µc µc µc µc 5. 6 µc
42 A Example: Find the equivalent capacity of the circuit shown in Fig. (a) C 2 C µ F 4.0 µ F C 3 C µ F 1.0 µ F B (a) A C 2 C µ F 4.0 µ F B (b) C µ F If we apply a voltage of 100V between points A and B what is the voltage across capacitor C2? A C 1 C µ F 3.0 µ F B (c) Ans: 25V A B (d) 7.0 µ F
43 1. (C eq ) a > (C eq ) b = (C eq ) c > (C eq ) d Rank in order, from largest to smallest, the equivalent capacitance (C eq ) a to (C eq ) d of circuits a to d. 2. (C eq ) b > (C eq ) a = (C eq ) d > (C eq ) c 3. (C eq ) c > (C eq ) a = (C eq ) d > (C eq ) b 4. (C eq ) d > (C eq ) b = (C eq ) c > (C eq ) a 5. (C eq ) d > (C eq ) b > (C eq ) a > (C eq ) c
44 1. (C eq ) a > (C eq ) b = (C eq ) c > (C eq ) d Rank in order, from largest to smallest, the equivalent capacitance (C eq ) a to (C eq ) d of circuits a to d. 2. (C eq ) b > (C eq ) a = (C eq ) d > (C eq ) c 3. (C eq ) c > (C eq ) a = (C eq ) d > (C eq ) b 4. (C eq ) d > (C eq ) b = (C eq ) c > (C eq ) a 5. (C eq ) d > (C eq ) b > (C eq ) a > (C eq ) c C=5 C=3+3=6 C=1/(1/3+1/3)=3/2 C=3+1/(1/4+1/4)=5
45 What is the equivalent capacitance, C eq, of the combination below? 1) C eq = 3/2 C 2) C eq = 2/3 C 3) C eq = 3 C 4) C eq = 1/3 C 5) C eq = 1/2 C o C eq C C C o
46 What is the equivalent capacitance, C eq, of the combination below? 1) C eq = 3/2 C 2) C eq = 2/3 C 3) C eq = 3 C 4) C eq = 1/3 C 5) C eq = 1/2 C The 2 equal capacitors in series add up as inverses, giving 1/2 C. These are parallel to the first one, which add up directly. Thus, the total o C eq C C C equivalent capacitance is 3/2 C. o
47 How does the voltage V 1 across the first capacitor (C 1 ) compare to the voltage V 2 across the second capacitor (C 2 )? 1) V 1 = V 2 2) V 1 > V 2 3) V 1 < V 2 4) all voltages are zero C 2 = 1.0 µf 10 V C 1 = 1.0 µf C 3 = 1.0 µf
48 How does the voltage V 1 across the first capacitor (C 1 ) compare to the voltage V 2 across the second capacitor (C 2 )? 1) V 1 = V 2 2) V 1 > V 2 3) V 1 < V 2 4) all voltages are zero The voltage across C 1 is 10 V. The combined capacitors C 2 +C 3 are parallel to C 1. The voltage across C 2 +C 3 is also 10 V. Since C 2 and C 3 are in series, their voltages add. Thus the voltage across C 2 and C 3 each has to be 5 V, which is less than V V C 2 = 1.0 µf C 1 = 1.0 µf C 3 = 1.0 µf
49 How does the charge Q 1 on the first capacitor (C 1 ) compare to the charge Q 2 on the second capacitor (C 2 )? 1) Q 1 = Q 2 2) Q 1 > Q 2 3) Q 1 < Q 2 4) all charges are zero C 2 = 1.0 µf 10 V C 1 = 1.0 µf C 3 = 1.0 µf
50 How does the charge Q 1 on the first capacitor (C 1 ) compare to the charge Q 2 on the second capacitor (C 2 )? 1) Q 1 = Q 2 2) Q 1 > Q 2 3) Q 1 < Q 2 4) all charges are zero We already know that the voltage across C 1 is 10 V C 2 = 1.0 µf and the voltage across C 2 and C 3 each is 5 V. Since Q = CV and C is the same for all the capacitors, then since V 1 > V 2 therefore Q 1 > Q V C 1 = 1.0 µf C 3 = 1.0 µf
51 Initially the switch is in position A and capacitors C2 and C3 are uncharged. Then the switch is flipped to position B. Afterward, what are the charge on and the potential difference across each capacitor? Ans: Q1 = 0.83 mc, Q2 = Q3 = 0.63 mc V1 = 55 V, V2 = 34 V, V3 = 21 V
52 You reposition the two plates of a capacitor so that the capacitance doubles. If the charges +Q and Q on the two plates are kept constant in this process, what happens to the energy stored in the capacitor? 1. increases by a factor of 4 2. increases by a factor of 2 3. the stored energy is unchanged 4. decreases by a factor of 2 5. decreases by a factor of 4
53 You reposition the two plates of a capacitor so that the capacitance doubles. If the charges +Q and Q on the two plates are kept constant in this process, what happens to the energy stored in the capacitor? 1. increases by a factor of 4 2. increases by a factor of 2 3. the stored energy is unchanged 4. decreases by a factor of 2 5. decreases by a factor of 4
54 You slide a slab of dielectric between the plates of a capacitor. As you do this, the charges on the plates remain constant. What effect does this have on the energy stored in the capacitor? 1. the stored energy increases 2. the stored energy is unchanged 3. the stored energy decreases 4. not enough information given to decide +Q Q +Q Q
55 You slide a slab of dielectric between the plates of a capacitor. As you do this, the charges on the plates remain constant. What effect does this have on the energy stored in the capacitor? 1. the stored energy increases 2. the stored energy is unchanged 3. the stored energy decreases 4. not enough information given to decide +Q Q +Q Q
56 Two 5.0cmdiameter metal disks separated by a 0.50mmthick piece of Pyrex glass are charged to a potential difference of 1000 V. What are (a) the surface charge density on the disks, and (b) the surface charge density on the glass? Look up in table 30.1, for Pyrex glass κ = 4.7 Ans: (a) 8.3 x 105 C/m 2 (b) 1.4 x 105 C/m 2
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