Ch16-19 lecture
Resistors in parallel
-Current splits into parallel branches so they add - same voltage drop in each branch b/c all pathways start and end at a common point I1 = I2 + I3 V = V1 = V2 -Reciprocal resistances add Requiv = (R1*R2)/(R1+R2 ) -Equivalent resistance decreases as more resistors are added in parallel -Sum of the currents going into each division must equal the total current going out 1/R=1/R₁+1/R₂+1/R₃+... V=V₁=V₂=... Each bulb is under the terminal voltage of the battery Same potential difference but different currents.
Electric current (I)
-potential difference or voltage/resistance i=(V/R) Electric current is the rate of flow of charge through a conductor. I=Delta Q/Delta t Q=charge Current I=Amperes (A) measured 1A=1C/sec Electric current goes from positive to negative conventional current : from negative to positive direction of current is direction of charge flow I=1000x1.6E-19)/1E-6sec = 1.6E-10A the direction is opposite in current is in opposite direction of electron flow Direction of the electric current is the direction of positive charge flow.
In the circuit given in figure, what would be the current on resistors and charge on the capacitor after leaving the switch on for a long time?(a)1 A, 3 mC. (b) 1 A, 9 mC. (c) 0 A, 0 mC. (d) 0 A, 9mC.
0 A, 9mC.
The change in the potential energy of an electron is (e=1.6E-19C) moved through a 100V potential difference is:
100V
Two 2ohm resistors are connected in parallel and the combination is connected in a series to a 11ohm resistor. If a 12 V potential difference is applied to this combination of three resistors, The potential difference between a 2 ohm resistor is other 2V 12V 1V
1V
parallel plate capacitor
2 plates of conductive material separated by small distance 1 plate positive 1 plate same negative charge -separation of charge creates electric field : E=(1/k)(Q/Ae0) A= plate area d=Plate seperation Sigma o=Q/A charge density (charge per unit area can calculate electric field E=sigma/epislon knot can calculate potential difference V=Ed=(sigma/epislon knot)d=(Qd)/epsilon not *A Q=CV C=Q/V=epsilon A/d C capacitance Capacitance depends only on geometrical parameters C=epsilon A/d IF YOu increase area and decrease d then you have greater capacity If you decrease A and Increase d then you have smaller capacity value For example you have a coffee can smaller one and larger one. Same height.. Small inside of large Now, if you dump positive Q charge into it. No one will gain negative Q charge and works as a cylinder capacitor can have spherical capacitor too.
Another example.
3-5-3
Electric charge
: Q or q Unit of electric charge is C Coulomb Coulomb is a basic SI Unit Electronic charge (e)= 1.6x10^-19 C will use micro, nano, or pico Coulombs
Using V=k(Q/r) And the more than one point charge equation.
A= 0 B=0 The electric field at B= cannot be 0
Energy stored in a parallel plate capacitor
Ad= Volume of the capacitor PE/Ad=1/2epsilon not E^2 Gives lower case u energy density u=1/2 epsilon not E^2 Electric energy is in the electric field
Last property of electric charge introduced Electric Charge is Quantized
An object is negatively charged when it has an excess of electrons, and positively charged when it has less than its normal number of electrons. The net charge on any object is a whole number times ±e or -e. That is, charge is quantized. Quantized: If you take any charge Q, +/-ne (n=integer) A decimal of an electric charge is not possible.
Kirchhoffs Rules
At any junction point, the sum of currents entering the junction must be equal to the sum of currents leaving the junction. For a closed circuit loop, sum of the potential differences must be zero.
Capacitor one has potential energy U1 when the potential difference is V1 while the capacitor two has potential energy U2 when the potential difference is V2. If U1=2U2 and V1=2V2 what must be the relationship between their capacitances? C1=C2/4 C1=C2/2 C1=2C2 C1=4C2
C1=C2/2
induction
Charging by conduction: "contact charging" two objects end up with the same sign of charge when a positively charged object touches an uncharged object. Charging by induction: Positive charge at one end and negative charge at the other end. No net charge created but the charges have been separated. Rubbed to charge
Alternating current (AC)
Current from a battery flows steadily in one direction (direct current, DC) goes in one direction from positive to negative Energy is proportional to the area of I^2 against time plot. Area is the energy A AC source: Current from a power plant varies sinusoidally (alternating current, AC). Changing like sin
The Watt-hour is a unit of electric power Electric bulbs Electric field Electric Energy
Electric Energy
Conductors and Insulators
Electric conductors are those materials in which many electrons are relatively free to move, whereas electric insulators or nonconductors are those in which very few electrons are free to move. Semiconductors are the intermediate category where although they are insulators they conduct electricity slightly (sillicon and Germanium). In insulating materials: electrons are bound tightly to the nuclei In good metal conductors: Electrons are bound freely within the metal although they cannot leave easily. This is often referred to ask free electrons or conduction electrons. Summation of conductors: When a positively charged object is brought close to or touches a conductor, the free electrons in the conductor are attracted by this positively charged object and move quickly toward it. If a negatively charged object is brought close to the conductor, the free electrons in the conductor move swiftly away from it. In a semiconductor, there are many fewer free electrons, and in an insulator, almost none. CONDUCTORS: any metals INSULATORS: rubber, wood, ceramic, plastic Conductor: Have free electrons; permit free charge flow; can be charged by induction. Induction is not the plastic rod/cloth example Insulators: opposite of conductors. Do not have free electrons (all electrons are bound); Do not permit free charge flow. Cannot be charged by induction (rub to charge). Coulomb's Law: Under the properties of electric charges. Opposite charges attract. then there is a force between the charges and is given by this law. Q1& Q2 is separated by r. gives force. epsilon knot= permittivity of free space. epsilon knot= 8.5x10^-12
Which of the following is a vector? Electric potential Electric charge Electric field Capacitance
Electric field
Electric potential
Electric potential: electrical potential of a +1 charge electric potential= V Electric potential= PE or U V=(PE)/q q=charge In electric field: downward. from positive charges to negative charges qEd=PE-- potential energy difference a and b electric potential energy PEb-PEa=-qEd electric potential Vb-Va=Vba PE/q=V Va=PEa/q Vb=PEb/q Vb-Va= PEb-PEa)/q =-2Ed)/2 =-Ed ---- Delta V
A resistor carrying a current I
Electric power (P): Power is the energy transformed by a device per unit time. When a Delta q charge passes through R, the work done, Delta W P=E/t Delta U=potential energy P=VI=V^2/R=I^2R
Example of magnitude of the Coulomb force
I have 3 charges sitting on the x-axis.
Example of property of Electric charge: Quantization
I have 4 objects and four different charges on those objects. measure the charges on the four objects using electroscope. Given multiple measurements Without going back to device a simple test can be done to see if measurements are correct. to see if they obey quantization of electric charge e=1.6x10^-19 Can divide give charge be e to see if I get an integer.
when you insert a dielectric into a capacitor
Net E=E0-Einduced V-Enet*d Enet<E0 Enet is smaller that original V<V0 Net is smaller by inserting a dielectric inside a capacitor electric field goes down as well as potential of walls of capacitor New Capacitance C=K(epsilon not A/d)=Epsilon (A/d) C=kC0 Epsilon=permittivity of the dielectric K=dielectric constant epsilon=k epsilon not C>C0 Capacitance can be increased by inserting a dielectric into the capacitor
Calculating negative potential energy
PE=(k(-2E-6)(2E-6)_/.5. negative positive and positive then PE positive
potential energy (PE) of point charges
PE=Vq=(kQ1/r)Q2 PE=k(Q1Q2/r)----for one pair for 3: Look at slide. add equations. PE of four charges: r12+r13+r23+r14+r24+r34
Series and Parallel capacitors
Parallel capacitors : same potential difference and different charge Series Capacitors: Same current (charge) but different potential differences for each
1.The equivalent resistant of a network of three 2 W resistors can be: (a) 0.33 W. (b) 1.5 W. (c) 1.33 W. (d) 2 W. (e) none. 2. Four resistors, 1 W, 10 W, 100 W, and 1000 W, are connected in parallel and 10 V potential difference is maintained across the combination. Mark the correct statement/statements. (a) The equivalent resistance of the circuit is closely equal to 1000 W. (b) All four resistors carry the same current. (c) The potential difference across the 100 W resistor is less than 10 V. (d) The current through the 100 W resistor is 1 A. (e) The current through the 10 W resistor is 1 A.3. What would be your answer if the resistors are Series? Two 4 W resistors are connected in parallel and the combination is connected in series with a 8W resistor. If a 10 V potential difference is applied to this circuit, the potential difference between the ends of 8 W resistor is (a) 10 V. (b) 8 V. (c) 4 V. (d) 6 V. (e) 2 V. (f) other.
SL - 2 c, e (abc), b.
The circuit shown at right is made using identical bulbs an ideal battery. a. if the resistance of the bulb is 4 ohms and the battery is 12V find the current in each resistor b. Is the bulb A brighter, dimmer or same brightness as bulb B? c. Is the current through bulb D greater than, less than or equal to the current through F? d. Is the current through bulb B greater than, less than, or equal to the current through bulb F? e. Bulb F is unscrewed from its socket, opening the circuit. Does bulb B Become brighter, dimmer, or stay the same brightness?
SL - 3 (a) 0.66 AC, 1.8 B, 1.2 F, 0.6 DE
If the potential and electric field at point A, respectively, are VAand EA while those at point B, again respectively, are VB and EB find the correct statement. (a)VA = VB and EA = EB (b)VA > VB and EA > EB (c) VA < VB and EA < EB (d)VA > VB and EA = EB (e)VA > VB and EA < EB An 100 V/m electric field is directed along the + y axis. If the potential at the origin is 20 V, what is potential at the point (2m, 3m) point? (a) 400 V. (b) 500 V. (c) zero. (d) - 100 V. (e) - 280 V. (f) other
SL - 5 d, e.
An electric iron draws a current current of 10 A when connected to a 120 V power source. Its resistance is (a) 10 W. (b) 120 W (c) 12 W (d) other. A certain wire has a resistance R. The resistance of another wire identical with first except having twice its diameter, is(a) R. (b) 2R. (c) R/2. (d) 4R. (e) R/4. (f) other. When a 60-W light bulb is operating on 120-V household line, what is the current in it? A) 0.5 A B) 120 J/C C) 2.0 A D) 1.414 A E) 40 J/C An unknown resistor dissipates 1.0 W when connected to a 2 V potential difference. When connected to a 1V potential difference, this resistor will dissipate: (a) 0.25 W. (b) 4.0 W. (c) 0.5 W. (d) None of these.
SL -1 c, e, a, a.
Six 12 W resistors are connected parallel and the combination is connected to a 15 V battery with negligible internal resistance. The equivalent resistance of the circuit is: (a) 6 W . (b) 12 W. (c) 60 W. (d) 2 W. (e) other.What would be your answer if the resistors are Series? Three identical resistors are connected in parallel to a battery. If the current of 9.0 A flows from the battery, how much current flows through any one of the resistors? (a) 4.5 A. (b) 0.5 A. (c) 3.0 A. (d) 1.5 A (e) other. Five 15 W resistors are connected series and the combination is connected to a 12 V battery with negligible internal resistance. The potential difference across any one of the resistor is (a) 3 V. (b) 12 V. (c) 4 V. (d) 5 V. (e) other. Two 2 W resistors are connected in parallel and this combination is connected in series with 3 W resistor. The resistors are connected to a 12-V battery. What is the potential difference between the ends of the 3 W resistor? (a) 9 V. (b) 6 V. (c) 15 V . (d) 3V. (e) Other.
SL -1 d (e), c, e, a.
Find the electric current through the resistor in each of the following circuits.
SL- 5 0.3, 0.15, 0.45, 0.75
What is the correct statement relating to the current in the figure? (a) Current in all three resistors must be zero . (b) Current in R1 and R2 must be zero. (c) Current in R2 and R3 must be zero. (d) Current in R1 and R3 must be zero. (e) Current in R1 must be zero. (f) Current in R2 must be zero. (g) Current in R3 must be zero 2.What is the potential at point A if R1 = R3 ? (a) Zero. (b) - 12 V. (c) + 12 V. (d) - 6 V. (e) +6 V. (f) other.
SL- 5 f, d.
Which of the followings do not affect the capacitance of a parallel plate capacitor ? (a)Area of plates. (b) Separations of the plates. (c)Material between plates. (d) Charge on the plates .(e) Potential difference of plates. (f) Energy stored in the capacitor. A parallel plate capacitor stores a charge Q = 4 nC when connected to a battery of V = 10 V. The capacitance of the capacitor must be; (a) 0.4 F. (b) 0.4 uF. (c) 40 nF. (d) 40 F. (e) other. The energy stored by a capacitor of capacitance C, when a voltage V is applied and a charge Q is stored on it, is not given by: (a) Q2C/2. (b) QV/2. (c) CV/2. (d) QC/2. (e) CV2/2.
SL- 6 (def), e, (acd).
1.The equivalent capacitance of a network of three 2 mF capacitors can be: (a) 0.33 mF. (b) 1.5 mF. (c) 3 mF. (d) 2 mF. (e) none.2. Using 1 mF, 2 mF, and 3 mF capacitors in combination, the least capacitance possible is (a) 1 mF. (b) 0.55mF. (c) 5 mF. (d) 0.33 mF. 3.1 uF, 2 uF, and 3 uF capacitors are connected in parallel, the combination being connected across a 9.0 V battery. The capacitor with the greatest charge on it is (a) 1mF. (b) 2 mF. (c) 3 mF. (d) neither, they all have the same Q What would be your answer if they are Series?
SL- 6 c, b, c(d).
The force on a positron (e = 1.6 x 10-19 C) in an electric field of 200 N/C is ( a) 3.2x 10-17 N to the direction of electric field. (b) 8.0 x 10-22 N against the direction of electric field. (c) 8.0 x 10-22 N to the direction of the electric field. (d) 3.2 x 10-17 N against the direction of electric field. An electron traveling west enters a region where the electric field is uniform and points north. The electron:( a) Veers south. (b) veers north. (c) slows down. (d) speeds up. The earth posses a net upward electric field near its surface. This is consistent with earth having (a) zero net charge. (b) negative charge. (c) positive charge. (d) magnetic field.
SL- 7 a, a, c.
To be safe during a lightning storm, it is best to be (a) in the middle of a grassy meadow. (b) Inside a metal car. (c) Under a tall tree. (d) Inside a wooden building. (e) on a metal observation towers. Five equal positive charges are uniformly spaced along the x axis. The electric field at point P (on y axis) must be (a) Zero. (b) to + X direction. (c) to - X direction. (d) to - Y direction. (e) to + y direction. (f) to other direction.
SL- 8, b, e.
Which of the followings are not vectors; (a)Electric potential. (b) Electric field. (c) Coulomb force. (d) Electric potential energy. (e) Equipotential lines. (f) Electric field lines. The unit of electric potential is volt (V). One V is same as (a)J. (b) J/kg. (c) J/C. (d) C/m. An electric field is directed from floor to ceiling. Mark the correct statement/statements. (a) Electric potential must be greatest in the ceiling. (b) Equipotential lines must be from floor to ceiling. (c) Equipotential lines must be horizontal. (d) Electric potential decreases in downward direction.
SL-1 (ade), c, c,
Consider the following statements for a neutral object. I. No negative charge in it. II. No positive charge in it. III. The net charge in it must be zero.Correct statement/statements must be: (a) I & II. (b) III only. (c) All of these. (d) other. Two electrically neutral materials are rubbed together. One acquires a net charge of - 6.4 x 10-10 C. The other must (a) have lost 6.4 x 10-10 electrons. (b) have gained 6.4 x 10-10 electrons. (c) have lost 6.4 x 10-10 protons. (d) lost other number of electrons. If a positively charged rod is brought near one end of an uncharged metal bar, the end of the metal bar closer to the charged rod will be charged (a) positive. (b) negative. (c) neutral. (d) both positive and negative.
SL-1 b, d, b.
Four point charges are positioned in a rectangle as shown in the figure. The electric field is zero at: (a) all the five points A, B, C, D, and O. (b) none of the five points. (c) the points A, O and C only. (d) the points B, O and D only. (e) the point O only.
SL-10 e.
A 3 C charge is in an electric field. What happens to potential energy and potential if that charge is replaced by a 6 uC charge?(a)The electric potential doubles, but the electric potential energy stays the same. (b) The electric potential energy doubles, but the electric potential stays the same. (c) Both the electric potential and potential energy double. (d) Both the electric potential and potential energy stay the same. Chose the incorrect statement/statements: (a) An electron tends to go from region of low potential to region of high potential. (b) A proton tends to go from region of low potential to region of high potential. (c) The direction of electric field is from lower potential to higher potential. (d) Electric potential has direction.
SL-2 b, (bcd).
If a 1 A current flows for two minutes, how much charge has passed through this circuit? (a) 10 C. (b) 1200 C. (c) 120 C. (d) 100 C (d) other. A flow of 1000 electrons per second from east to west constitutes a current of (a) 1.6 X 10-17 A to west. (b) 1.6 X 10-17 A to east. (c) 1.6 X 10-21 A to west. (d) other value to east. If 120 V (Vrms) is in household electric line what is the peak voltage (Vo) of it? (a) 120 V. (b) 240 V. (c) 85 V. (d) 170 V. (e) other. When a 60 W bulb is connected to above house hold line, what is the peak current in it?(a) 0.5 A. (b) 2 A. (c) 0.35 A. (d) 0.7 A. (e) other.
SL-2 c, d, d, d.
Q1 = - 0.10 mC is located at the origin. Q2 = - 0.20 mC is located on the positive x axis at x = 1.0 m. Which of theFollowing is true of the force on Q1 due Q2?(a)It is attractive and directed in the +x direction.(b)It is attractive and directed in the - x direction.(c)It is repulsive and directed in the +x direction.(d)It is repulsive and directed in the - x direction. Two charged particles attract each other with a force F. If the charge on one of them is doubled and the distance between them is also doubled, then the force will be(a) F. (b) 2F. (c) F/2. (d) 4F. (e) F/4. (f) other.
SL-2 d, c.
Three table tennis balls are suspended from thin threads. It isfound that ball 1 and ball 2 repel each other and that ball 2 and 3 attract each other. From this we can conclude that: (a) 1 and 3 carry charge of equal sign. (b) 1 and 3 carry charge of opposite sign. (c) all three carry the charge of the same sign. (d) one of the object carry no charge. (e) need more information to determine the sign of the charges.What would be your answer if the balls are aluminum in above question?
SL-3 a, e.
Four point charges are positioned in a rectangle as shown in the figure. The potential is zero at: (a) all the five points A, B, C, D, and O. (b) none of the five points. (c) the points A, O and C only. (d) the points B, O and D only. (e) the point O only.
SL-3 a.
Along the direction of the electric field the potential ________. If a free electron is placed in an electric field, it will accelerate towards the ________ potential while a free proton will accelerate towards the ______ potential. Which of the followings fills the three missing words in correct order? (a) increases, high, low. (b) decreases, low, high. (c) increases, low, high. (d) decreases, high, low. In a certain region of space the electric potential decreases uniformly from north to south and does not vary in any other direction. The electric field: (a) points north and varies with position. (b) points north and does not vary with position. (c) points south and varies with position. (d) points south and does not vary with position If 500 eV of work are required to carry a 40 electrons from one point to another, the potential difference between these two points is: (a) 7.8 x 1019 V. (b) 0.08 V. (c) 12.5 V. (d) other.
SL-3 d, d, c.
If the electric field is 10 V/m in the positive y direction, what is the potential difference between the origin, (0, 0), and the point(2m, 3m )?(a) 10 V. (b) 20 V. (c) 30 V. (d) other. 3. If the electric field is 10 V/m in a direction that makes 45o to the direction of +x axis, what is the potential difference between the points (0, 2m) and (2m, 0)? (a) zero. (b) 20 V. (c) 20 V (d) 57 V. (e) other. An 200 V/m electric field is directed along the + x axis. If the potential at the origin is 500 V, what is potential at the point ( 2m, 0) point? (a) 400 V. (b) 500 V. (c) zero. (d) - 100 V. (e) - 300 V. (f) other An 100 V/m electric field is directed along the + x axis. If the potential at the origin is 300 V, what is potential at the point ( -2m, 0) point? (a) 400 V. (b) 500 V. (c) zero. (d) - 100 V. (e) - 300 V. (f) otherWhat is the potential at (-2m, -3m) point?
SL-4 c, a, f, b
Tworesistors, R1 and R2, are connected in parallel and the combination is connected in series with resistor R3. If R1 = R2 = R3 and a potential difference is applied to this circuit, the power dissipated is (a) greatest R1. (b) greatest in R2. (c) greatest in R3. (d) The same in all three resistors. Consider the circuit given in above MCQ. If R2= 2R1, R3 = 3R1, and the current in R1 resistor is I1, the current in R3 must be:(a) I1. (b) 2I1. (c) 3I1. (d) 1.5 I1. (e) 0.66 I1. (f) other. For the given circuit, which of the following equation is correct? (a) 30I1 + 21I2 = 80. (b) 30I1 - 41I3 = 45. (c) 30I1 +41I3 = 80. (d) 41I3 + 21I2 = 125 . (e) I don't believe any of these are correct.
SL-4 c, d, d.
When previously uncharged object is charged to - 1.6 X 10-6 C (e = 1.6 x 10-19 C) (a) it has an excess of positrons. (b) one million electrons have been removed from it. (c) 10 13 electrons have been added to it. (d) no choice is true. A small metal ball hangs from the ceiling by an insulating thread. The ball is attracted to a positively charged rod held near the ball. The charge of the ball must be(a) positive. (b) negative. (c) neutral.(d) positive or neutral. (e) negative or neutral.
SL-4 c, e.
For a positive charge moving in the direction of the electric field, a). its potential energy increases and its electric potential increases. b). its potential energy increases and its electric potential decreases. c). its potential energy decreases and its electric potential increases. d). its potential energy decreases and its electric potential decreases. e). its potential energy and electric potential remain constant.A positive charge is moved from one point to another along an equipotential surface. The work required a). is positive. b). is negative. c). depends on the sign of potential. d). is zero. e). depends on the magnitude of potential. The electric potential measured at a point equidistant from two particles that have charges equal in magnitude but of opposite sign is a). larger than zero. b). smaller than zero. c). equal to zero. d). equal to the average of the two distances times the charge. e). equal to the net electric field.
SL-4 d, d, c.
The force on charge 4 mC is (a)in the + X direction. (b) in the - X direction. (c)zero. (d) in the +y direction. (e) in the - y direction. (f) other.
SL-5 e.
The direction of the electric field halfway between an electron and a proton is______. A)undefined since the electric field is zero B) toward the proton C) toward the electron D) perpendicular to the line from the electron to the proton E) cannot be determined In a region where the electric field is uniform, the electric field lines ________ A) are parallel to one another but the spacing can change from point to point. B) are parallel to one another and equally spaced. C) diverge at a constant rate. D) converge if a negative charge is involved. E) cannot be determined.
SL-6 c, b
A 2uF capacitor is fully charged by connecting across a 1.5 Vbattery. Then the battery is replaced by a 1 uF capacitor. What is the final charge in the 2 uF capacitor? (a) 3 C. (b) 1 C. (c) 2 C. (d) 1.5 C. (e) other. If the voltage between the plates of a parallel plate capacitor is doubled, the capacitance of the capacitor (a) is helved. (b) remains the same. (c) is doubled. (d) quadruples. If the charge on a capacitor is doubled, its stored energy is (a) halved. (b) doubled. (c) quartered. (d) changed by other factor.
SL-7 c, b, d.
If the voltage between the plates of a parallel plate capacitor is doubled, (a) the charge in the capacitor is halved. (b) the capacitance in the capacitor is helved. (c) the charge in the capacitor is doubled. (d) the capacitance in the capacitor is doubled Two isolated metallic spheres each have a net charge Q uniformly distributed over their surfaces. One sphere has a radius R and distributed over their surfaces. One sphere has a radius R and the other has r ( r < R). Which charge distribution stores more electric energy? a. The sphere of radius R. b. The sphere with radius r. a. The sphere of radius R. b. The sphere with radius r. c. Both have the same energy. d. Need more information. c. Both have the same energy. d. Need more information. e. I don't know.
SL-8. C, b.
A hollow conductor is positively charged. A small uncharged metal ball is lowered by a silk thread through a small opening in the top of the conductor and allowed to touch its inner surface. After the ball is removed, it will have: (a) a positive charge. (b) a negative charge. (c) no appreciable charge. (d) a charge whose sign depends on what part of the inner surface it touched. Two identical tiny spheres have the same electric charge. If the charge on each of them doubled, and their separation is also doubled, the force each exerts on the other will be (a) double. (b) unchanged. (c) four times larger. (d) changed by other factor. The electric field lines of a negative charge (a) circle clockwise around. (b) circle counter-clockwise around. (c) radiate toward the charge. (d) radiate outward from the charge.
SL-9 c, b, c.
The plates of a parallel plate capacitor are maintained with constantvoltage by a battery as they are pulled apart. During this process, theamount of charge on the plates must (a) increases. (b) decreases. (c) remains constant. (d) Need more information to answer this question. The plates of a parallel plate capacitor are maintained with constantvoltage by a battery as they are pulled apart. What happens to the strength of the electric field during this process?(a) increase. (b) decreases. (c) remain constant.(d) Need more information to determine this.
SL-9. b, b.
Energy (PE) stored in a Capacitor
Stored energy is equal to work done to store that charge. Energy stored = work done to charge capacitor A charged capacitor stores electric energy; the energy stored ie equal to the work done to charge the capacitor PE= 1/2 Q^2/C=1/2QV=1/2CV^2 A capacitor stores 4 J of energy when connected to 3 V battery.If the voltage of the battery is 6 V what is the energy stored in the capacitor PE2/PE1=6^2/3^2=4/1 PE2=4PE1=16 j Cant use PE=1/2QV because there are changes and no constant and C is a constant and is easier to solve
configuration
The capacitor continued There is a potential difference between the two plates. adding more charges then V will increase Q- Charge is directly proportional to V-potential difference Q=CV----C is constant C- Constant Capacitance (geometry thing) Q= Coulomb (C) V= Volts (V) C= Farad (F) micro, nano, or pico V=Ed, d plate separation, E-Electric field The capacitor is two even lines The battery is two different lengths and a switch is a open line connect capacitor to a battery to supply voltage V Q=CV Q=5E-6Fx1.5V=7.5E-6 FV=C 7.5 micro C
The Electric Field E
The force exerted on a tiny +1 charge (test charge) is defined as the electric field. q= test charge Electric field is a force and a vector E is a vector If Q is positive: The force is opposite direction away from electric charge. If Q is negative: the direction of the force is towards the charge Unit; N/C
properties of Q or (q)
The law of conservation of charge states that: the net amount of electric charge produced in any process is zero; no net electric charge can be created or destroyed. Plastic tube: Rub tube with silk cloth to make a charge. Initially the charge is neutral. Net charge= 0 Qnetplastic=0 Qsilk=0 Once rubbed together The plastic is: negative The silk is: positive charged What happened is that the silk cloth repulsed electrons from the plastic tube so plastic tube lost electron and silk gained electrons. Qnet silk= -5e Qnet plastic= +5e TotalQnet=0 Initial and final net charge is 0 Cannot create or destroy charges but can transfer charges from one object to another. Electric charges are conserved The term neutral means net charge =0
how the electric currents and charge and relationships there
The source of electric current: battery How battery produces current and charge 2 metals in sulfuric acid Carbon and zinc electrodes Electricity could be created if dissimilar metals were connected by a conductive solution called an electrolyte. Chemical reaction between metals and reaction. They gain positive and negative charges A battery transforms chemical energy into electrical energy and maintains a potential difference between its terminals. Chemical energy into electrical energy. Deposit positive into one electrode and negative into the other electrode and it creates a potential difference to draw current Potential difference as a result of the charge separation.
Properties of charges
There are two kinds of charges positive and negative charges. There are two types of electric charge: positive and negative. One repulsive force and the other is attractive force. Unlike charges attract and like charges repel. Smallest possible charge is positive or negative e Electronic charge (e)= 1.6x10^-19 C The negative e is an electron Positive e is a positron
Units of V
V: volts is the unit of potential potential difference: Vba=DeltaV=EdeltaX Delta V= Volts E=N/C X= Meters E=DeltaV/DeltaX Volt/meter V/m=N/C. are the same Electron Volt: eV is a unit of energy (or work) place an electron in higher potential side. goes to higher potential. calculate work using potential energy change= Charge x potential difference. ex1V1=1eV On white board PE=qV 1eV=1.6E-19 C * 1V -1.6E-19CV== Joules =1.6E-19J Electric potential (V) due to a point charge: E and V at point A? E=k(Q/r^2) k=9E9 Nm^2/C^2 V=K(Q/r) at infinity V=0 V is a scalar quantity has no direction only magnitude. more than one point charge V=SumV V=-Ed not valued because it is not constant E not constant
Ohm's law: Resistance and Resistors
Voltage proportional to Current V=IR is Ohms law R=Resistance, has units of Volt/amp= Ohms It is a constant for any given V=IR Q=CV C=constant R=constant no kilo ohms just written as k when in 1000 is you see m it is x10^-6 ohms m=mega ohms A hair dryer draws 9.0 A current when plugged into 120 V power outlet. What is the resistance of hair dryer? V=IR 120v=9.0A* R R=120/9= 13.3ohms
In DC circuits DC: Direct current circuits
Watts V=IR ohms law EMF: electric motor force. Not a force but a potential difference and terminal voltage of a battery: The potential difference Vab. The large E near the battery is the electric motor force. The term r here is the internal resistance of the battery. Battery is nearly constant voltage source, but does have a small internal resistance (r), which reduces the actual voltage from the ideal emf (e): Vab=epsilon-Ir Ir= potential drop in the resistor EMF is the terminal potential when there is no current coming from the battery. If no current from battery then Vab=epsilon (EMF) r is nearly zero for a new battery r gets larger through prolonged use of the battery. thats why you have to change batteries
Dielectrics
are insulators that have randomly oriented dipoles The dipoles can rotate but cannot move. They are bound. Consider a capacitor with capacitance C0 , charge Qo and potential difference V0 - at the beginning Q0=C0V0 V0=E0d Insert the dielectric the dipoles are able to rotate and positive towards negative and negative towards positive. E field inside is reduced
(a) What is the current through electric heater?(b) What is the total current through the fuse (or circuit breaker)?(c) If the fuse is rated as 30 A, can You replace 100 W bulb by a 50 W bulb?(d) How about a 200 W bulb for (c)? A household electric line is equipped with a 10A fuse. The maximum number of 60-W 120-V light bulbs those can be powered on this circuit is: (a) 10. (b) 20. (c) 25. (d) 40.
b=20
Example:
find the electric field of -2 charge
if -2 on x axis is
in positive x direction
Electric field at two levels
level a is higher level b is lower Vba=-Ed Place the charge q at higher potential: then moves down PEb-PEa=-qED=qVba Potential difference The example of electric field on X/Y axis and the electric field E=100N/C In X,Y plane Point A -3,2 B= 4,2 C=4,-1 Now calculate the potential differences What are the Vba Vcb Vca? *look at differences in X coordinates. ** Electric field is in the X axis direction so you are only looking at x coordinance Between A&b the difference is 7 so as a result the Vba= -700Nm/C--- the d value was zero due to no y component Vcb= have the same potential. 0 Equipotential line. dont know the potential but it is the same along that line. Equipotential lines are always 90 degrees to the electric field line. Vca= -700 The Equipotential lines always go from higher potential to the lower potential Potential difference = V=E delta x or V=E delta y If free proton and electron placed on Equipotential line. The free proton exerts force into the way of the electric field Free electron puts in opposite direction accelerates negative ***free positive charge particles goes from higher potential to lower potential Negative charge goes from lower potential to higher**
Electric field lines
paths of free positive point charges are electric field lines. The strongest field is in the middle. Walls on LT and RT. Neg RT and Pos LT. from post to neg in 3D case. Always from positive to negative
Resistivity (p)
property of a substance; a quantitative measure how much charge flow a substance will permit resistance of a material changes Short and and long one The long one takes longer for charges to flow so more resistance. Greater length then greater resistance. Greater thickness: lower resistance R=rho (l/A) roh=ohm*meter= roh cu (gold) =10^-9 ohm*m gold is the best conductor ceramic: high roh = 10^12 good insulator Varies with temperature : Units:For any given conducting material, the resistivity increases with temperature As T increases greater Resistance Rt=R0(1+alpha(T-T0)
CHAPTER 17 Lecture 2 Capacitance (C)
the ability of a conductor to store energy in the form of electrically separated charges we have 2 metal plates next to each other and they are aligned parallel. the two plates have to be conductors. :metal because conductors have free charge and also the conductors are neutral meaning they have equal amount of positive and negative charges the net charge in each one is zero Labeled (1) & 2) plate dump some positive charge given to plate 1 and 2 If you do that then the configuration is all positive in first plate and zero in the second because of initial state Consider the effect of the positive charge: more negative on inside of plate 2 and more positive on outside of plate 2 (induction) Then, the positive charges of the second plate will be taken out when the surface is touched. Now that the second plat was touched there is no positive charge on the outside of plate 2 So you will have positive charges in 1st and negative charges in the second as a result of the induction. Induced negative charges in second plate. The capacitor stores charge Q
Magnitude of the Coulomb force
you can find this You can disregard the signs of charges and take the magnitude of the Q1 and magnitude of the Q2. You don't know direction but you know magnitude. to find the direction: positive charge and negative charge. attractive forces Positive and positive charge: Repulsive forces you can calculate the magnitude of the force and the direction you can get from the picture. They are in opposite directions Look at the nature of the charges to find the force.
Series vs. Parallel
•In a series process you can only follow one logical sequence: R=R1+R2+R3 •In a parallel process you can follow many different sequences 1/Req=1/R1+1/R2+1/R3 Parallel: multiple things plugged into a outlet Series: Christmas lights is one goes bad none of them work
