Properties of Electric Charge

Properties of Electric Charge
There are two kinds of electric charge. like charges repel unlike charges attract Electric charge is conserved. Positively charged particles are called protons. Uncharged particles are called neutrons. Negatively charged particles are called electrons.

Properties of Electric Charge, continued
Electric charge is quantized. That is, when an object is charged, its charge is always a multiple of a fundamental unit of charge. Charge is measured in coulombs (C). The fundamental unit of charge, e, is the magnitude of the charge of a single electron or proton. e = x 10–19 C

Transfer of Electric Charge
An electrical conductor is a material in which charges can move freely. An electrical insulator is a material in which charges cannot move freely.

Transfer of Electric Charge, continued
Insulators and conductors can be charged by contact. Conductors can be charged by induction. Induction is a process of charging a conductor by bringing it near another charged object and grounding the conductor.

Transfer of Electric Charge, continued
A surface charge can be induced on insulators by polarization. With polarization, the charges within individual molecules are realigned such that the molecule has a slight charge separation.

Electric Force Coulomb’s Law Two charges near one another exert a force on one another called the electric force. Coulomb’s law states that the electric force is propor-tional to the magnitude of each charge and inversely proportional to the square of the distance between them.

Coulomb’s Law, continued
Electric Force Coulomb’s Law, continued The resultant force on a charge is the vector sum of the individual forces on that charge. Adding forces this way is an example of the principle of superposition. When a body is in equilibrium, the net external force acting on that body is zero.

Electric Force Sample Problem The Superposition Principle Consider three point charges at the corners of a triangle, as shown at right, where q1 = 6.00  10–9 C, q2 = –2.00  10–9 C, and q3 = 5.00  10–9 C. Find the magnitude and direction of the resultant force on q3.

Sample Problem, continued
Electric Force Sample Problem, continued The Superposition Principle 1. Define the problem, and identify the known variables. Given: q1 =  10–9 C r2,1 = 3.00 m q2 = –2.00  10–9 C r3,2 = 4.00 m q3 =  10–9 C r3,1 = 5.00 m q = 37.0º Unknown: F3,tot = ? Diagram:

Electric Force Sample Problem, continued The Superposition Principle Tip: According to the superposition principle, the resultant force on the charge q3 is the vector sum of the forces exerted by q1 and q2 on q3. First, find the force exerted on q3 by each, and then add these two forces together vectorially to get the resultant force on q3. 2. Determine the direction of the forces by analyzing the charges. The force F3,1 is repulsive because q1 and q3 have the same sign. The force F3,2 is attractive because q2 and q3 have opposite signs.

Electric Force Sample Problem, continued The Superposition Principle 3. Calculate the magnitudes of the forces with Coulomb’s law.

Electric Force Sample Problem, continued The Superposition Principle 4. Find the x and y components of each force. At this point, the direction each component must be taken into account. F3,1: Fx = (F3,1)(cos 37.0º) = (1.08  10–8 N)(cos 37.0º) Fx = 8.63  10–9 N Fy = (F3,1)(sin 37.0º) = (1.08  10–8 N)(sin 37.0º) Fy = 6.50  10–9 N F3,2: Fx = –F3,2 = –5.62  10–9 N Fy = 0 N

Electric Force Sample Problem, continued The Superposition Principle 5. Calculate the magnitude of the total force acting in both directions. Fx,tot = 8.63  10–9 N – 5.62  10–9 N = 3.01  10–9 N Fy,tot = 6.50  10–9 N + 0 N = 6.50  10–9 N

Electric Force Sample Problem, continued The Superposition Principle 6. Use the Pythagorean theorem to find the magni-tude of the resultant force.

Electric Force Sample Problem, continued The Superposition Principle 7. Use a suitable trigonometric function to find the direction of the resultant force. In this case, you can use the inverse tangent function:

Coulomb’s Law, continued
Electric Force Coulomb’s Law, continued The Coulomb force is a field force. A field force is a force that is exerted by one object on another even though there is no physical contact between the two objects.

Electric Field Strength
The Electric Field Electric Field Strength An electric field is a region where an electric force on a test charge can be detected. The SI units of the electric field, E, are newtons per coulomb (N/C). The direction of the electric field vector, E, is in the direction of the electric force that would be exerted on a small positive test charge.

Electric Field Strength, continued
The Electric Field Electric Field Strength, continued Electric field strength depends on charge and distance. An electric field exists in the region around a charged object. Electric Field Strength Due to a Point Charge

Sample Problem Electric Field Strength
The Electric Field Sample Problem Electric Field Strength A charge q1 = µC is at the origin, and a charge q2 = –5.00 µC is on the x-axis m from the origin, as shown at right. Find the electric field strength at point P,which is on the y-axis m from the origin.

The Electric Field Sample Problem, continued Electric Field Strength 1. Define the problem, and identify the known variables. Given: q1 = µC = 7.00  10–6 C r1 = m q2 = –5.00 µC = –5.00  10–6 C r2 = m = 53.1º Unknown: E at P (y = m) Tip: Apply the principle of superposition. You must first calculate the electric field produced by each charge individually at point P and then add these fields together as vectors.

The Electric Field Sample Problem, continued Electric Field Strength 2. Calculate the electric field strength produced by each charge. Because we are finding the magnitude of the electric field, we can neglect the sign of each charge.

The Electric Field Sample Problem, continued Electric Field Strength 3. Analyze the signs of the charges. The field vector E1 at P due to q1 is directed vertically upward, as shown in the figure, because q1 is positive. Likewise, the field vector E2 at P due to q2 is directed toward q2 because q2 is negative.

The Electric Field Sample Problem, continued Electric Field Strength 4. Find the x and y components of each electric field vector. For E1: Ex,1 = 0 N/C Ey,1 = 3.93  105 N/C For E2: Ex,2= (1.80  105 N/C)(cos 53.1º) = 1.08  105 N/C Ey,1= (1.80  105 N/C)(sin 53.1º)= –1.44  105 N/C

The Electric Field Sample Problem, continued Electric Field Strength 5. Calculate the total electric field strength in both directions. Ex,tot = Ex,1 + Ex,2 = 0 N/C  105 N/C = 1.08  105 N/C Ey,tot = Ey,1 + Ey,2 = 3.93  105 N/C – 1.44  105 N/C = 2.49  105 N/C

The Electric Field Sample Problem, continued Electric Field Strength 6. Use the Pythagorean theorem to find the magnitude of the resultant electric field strength vector.

The Electric Field Sample Problem, continued Electric Field Strength 7. Use a suitable trigonometric function to find the direction of the resultant electric field strength vector. In this case, you can use the inverse tangent function:

The Electric Field Sample Problem, continued Electric Field Strength 8. Evaluate your answer. The electric field at point P is pointing away from the charge q1, as expected, because q1 is a positive charge and is larger than the negative charge q2.

The Electric Field Electric Field Lines The number of electric field lines is proportional to the electric field strength. Electric field lines are tangent to the electric field vector at any point.

Conductors in Electrostatic Equilibrium
The Electric Field Conductors in Electrostatic Equilibrium The electric field is zero everywhere inside the conductor. Any excess charge on an isolated conductor resides entirely on the conductor’s outer surface. The electric field just outside a charged conductor is perpendicular to the conductor’s surface. On an irregularly shaped conductor, charge tends to accumulate where the radius of curvature of the surface is smallest, that is, at sharp points.

Multiple Choice 1. In which way is the electric force similar to the gravitational force? A. Electric force is proportional to the mass of the object. B. Electric force is similar in strength to gravitational force. C. Electric force is both attractive and repulsive. D. Electric force decreases in strength as the distance between the charges increases.

Multiple Choice, continued
1. In which way is the electric force similar to the gravitational force? A. Electric force is proportional to the mass of the object. B. Electric force is similar in strength to gravitational force. C. Electric force is both attractive and repulsive. D. Electric force decreases in strength as the distance between the charges increases.

2. What must the charges be for A and B in the figure so that they produce the electric field lines shown? F. A and B must both be positive. G. A and B must both be negative. H. A must be negative, and B must be positive. J. A must be positive, and B must be negative.

3. Which activity does not produce the same results as the other three? A. sliding over a plastic-covered automobile seat B. walking across a woolen carpet C. scraping food from a metal bowl with a metal spoon D. brushing dry hair with a plastic comb

4. By how much does the electric force between two charges change when the distance between them is doubled?

Use the passage below to answer questions 5–6. A negatively charged object is brought close to the surface of a conductor, whose opposite side is then grounded. 5. What is this process of charging called? A. charging by contact B. charging by induction C. charging by conduction D. charging by polarization

Use the passage below to answer questions 5–6. A negatively charged object is brought close to the surface of a conductor, whose opposite side is then grounded. 6. What kind of charge is left on the conductor’s surface? F. neutral G. negative H. positive J. both positive and negative

Use the graph below to answer questions 7–10. The graph shows the electric field strength at different distances from the center of the charged conducting sphere of a Van de Graaff generator. 7. What is the electric field strength 2.0 m from the center of the conducting sphere? A. 0 N/C B. 5.0  102 N/C C. 5.0  103 N/C D. 7.2  103 N/C

Use the graph below to answer questions 7–10. The graph shows the electric field strength at different distances from the center of the charged conducting sphere of a Van de Graaff generator. 8. What is the strength of the electric field at the surface of the conducting sphere? F. 0 N/C G. 1.5  102 N/C H. 2.0  102 N/C J. 7.2  103 N/C

Use the graph below to answer questions 7–10. The graph shows the electric field strength at different distances from the center of the charged conducting sphere of a Van de Graaff generator. 9. What is the strength of the electric field inside the conducting sphere? A. 0 N/C B. 1.5  102 N/C C. 2.0  102 N/C D. 7.2  103 N/C

Use the graph below to answer questions 7–10. The graph shows the electric field strength at different distances from the center of the charged conducting sphere of a Van de Graaff generator. 10. What is the radius of the conducting sphere? F. 0.5 m G. 1.0 m H. 1.5 m J. 2.0 m

Short Response 11. Three identical charges (q = +5.0 mC) are along a circle with a radius of 2.0 m at angles of 30°, 150°, and 270°, as shown in the figure.What is the resultant electric field at the center?

Short Response, continued
11. Three identical charges (q = +5.0 mC) are along a circle with a radius of 2.0 m at angles of 30°, 150°, and 270°, as shown in the figure.What is the resultant electric field at the center? Answer: 0.0 N/C

12. If a suspended object is attracted to another object that is charged, can you conclude that the suspended object is charged? Briefly explain your answer.

12. If a suspended object is attracted to another object that is charged, can you conclude that the suspended object is charged? Briefly explain your answer. Answer: not necessarily; The suspended object might have a charge induced on it, but its overall charge could be neutral.

13. One gram of hydrogen contains 6.02  1023 atoms, each with one electron and one proton. Suppose that 1.00 g of hydrogen is separated into protons and electrons, that the protons are placed at Earth’s north pole, and that the electrons are placed at Earth’s south pole. Assuming the radius of Earth to be 6.38  106 m, what is the magnitude of the resulting compressional force on Earth?

13. One gram of hydrogen contains 6.02  1023 atoms, each with one electron and one proton. Suppose that 1.00 g of hydrogen is separated into protons and electrons, that the protons are placed at Earth’s north pole, and that the electrons are placed at Earth’s south pole. Assuming the radius of Earth to be 6.38  106 m, what is the magnitude of the resulting compressional force on Earth? Answer: 5.12  105 N

14. Air becomes a conductor when the electric field strength exceeds 3.0  106 N/C. Determine the maximum amount of charge that can be carried by a metal sphere 2.0 m in radius.

14. Air becomes a conductor when the electric field strength exceeds 3.0  106 N/C. Determine the maximum amount of charge that can be carried by a metal sphere 2.0 m in radius. Answer: 1.3  10–3 C

Extended Response Use the information below to answer questions 15–18.
A proton, which has a mass of  10–27 kg, accelerates from rest in a uniform electric field of 640 N/C. At some time later, its speed is 1.2  106 m/s. 15. What is the magnitude of the acceleration of the proton?

Extended Response, continued
Use the information below to answer questions 15–18. A proton, which has a mass of  10–27 kg, accelerates from rest in a uniform electric field of 640 N/C. At some time later, its speed is 1.2  106 m/s. 15. What is the magnitude of the acceleration of the proton? Answer: 6.1  1010 m/s2

Use the information below to answer questions 15–18. A proton, which has a mass of  10–27 kg, accelerates from rest in a uniform electric field of 640 N/C. At some time later, its speed is 1.2  106 m/s. 16. How long does it take the proton to reach this speed?

Use the information below to answer questions 15–18. A proton, which has a mass of  10–27 kg, accelerates from rest in a uniform electric field of 640 N/C. At some time later, its speed is 1.2  106 m/s. 16. How long does it take the proton to reach this speed? Answer: 2.0  10–5 s

Use the information below to answer questions 15–18. A proton, which has a mass of  10–27 kg, accelerates from rest in a uniform electric field of 640 N/C. At some time later, its speed is 1.2  106 m/s. 17. How far has it moved in this time interval?

Use the information below to answer questions 15–18. A proton, which has a mass of  10–27 kg, accelerates from rest in a uniform electric field of 640 N/C. At some time later, its speed is 1.2  106 m/s. 17. How far has it moved in this time interval? Answer: 12 m

Use the information below to answer questions 15–18. A proton, which has a mass of  10–27 kg, accelerates from rest in a uniform electric field of 640 N/C. At some time later, its speed is 1.2  106 m/s. 18. What is its kinetic energy at the later time?

Use the information below to answer questions 15–18. A proton, which has a mass of  10–27 kg, accelerates from rest in a uniform electric field of 640 N/C. At some time later, its speed is 1.2  106 m/s. 18. What is its kinetic energy at the later time? Answer: 1.2  10–15 J

19. A student standing on a piece of insulating material places her hand on a Van de Graaff generator. She then turns on the generator. Shortly thereafter, her hairs stand on end. Explain how charge is or is not transferred in this situation, why the student is not shocked, and what causes her hairs to stand up after the generator is started.

19. (See previous slide for question.) Answer: The charge on the sphere of the Van de Graaff generator is transferred to the student by means of conduction. This charge remains on the student because she is insulated from the ground. As there is no path between the student and the generator and the student and the ground by which charge can escape, the student is not shocked. The accumulation of charges of the same sign on the strands of the student’s hair causes the strands to repel each other and so stand on end.

Electric Charge Charging By Induction

Transfer of Electric Charge

The Electric Field Electric Field Lines

Electrical Potential Energy
Electric Potential Electrical Potential Energy Electrical potential energy is potential energy associated with a charge due to its position in an electric field. Electrical potential energy is a component of mechanical energy. ME = KE + PEgrav + PEelastic + PEelectric

Electrical Potential Energy, continued
Electric Potential Electrical Potential Energy, continued Electrical potential energy can be associated with a charge in a uniform field. Electrical Potential Energy in a Uniform Electric Field PEelectric = –qEd electrical potential energy = –(charge)  (electric field strength)  (displacement from the reference point in the direction of the field)

Electric Potential Potential Difference Electric Potential equals the work that must be performed against electric forces to move a charge from a reference point to the point in question, divided by the charge. The electric potential associated with a charge is the electric energy divided by the charge:

Potential Difference, continued
Electric Potential Potential Difference, continued Potential Difference equals the work that must be performed against electric forces to move a charge between the two points in question, divided by the charge. Potential difference is a change in electric potential.

Electric Potential Potential Difference, continued The potential difference in a uniform field varies with the displacement from a reference point. Potential Difference in a Uniform Electric Field ∆V = –Ed potential difference = –(magnitude of the electric field  displacement)

Sample Problem Potential Energy and Potential Difference
Electric Potential Sample Problem Potential Energy and Potential Difference A charge moves a distance of 2.0 cm in the direction of a uniform electric field whose magnitude is 215 N/C.As the charge moves, its electrical potential energy decreases by 6.9  J. Find the charge on the moving particle. What is the potential difference between the two locations?

Electric Potential Sample Problem, continued Potential Energy and Potential Difference Given: ∆PEelectric = –6.9  10–19 J d = m E = 215 N/C Unknown: q = ? ∆V = ?

Electric Potential Sample Problem, continued Potential Energy and Potential Difference Use the equation for the change in electrical potential energy. PEelectric = –qEd Rearrange to solve for q, and insert values.

Electric Potential Sample Problem, continued Potential Energy and Potential Difference The potential difference is the magnitude of E times the displacement.

Electric Potential Potential Difference, continued At right, the electric poten-tial at point A depends on the charge at point B and the distance r. An electric potential exists at some point in an electric field regardless of whether there is a charge at that point.

Electric Potential Potential Difference, continued The reference point for potential difference near a point charge is often at infinity. Potential Difference Between a Point at Infinity and a Point Near a Point Charge The superposition principle can be used to calculate the electric potential for a group of charges.

Capacitors and Charge Storage
Capacitance Capacitors and Charge Storage A capacitor is a device that is used to store electrical potential energy. Capacitance is the ability of a conductor to store energy in the form of electrically separated charges. The SI units for capacitance is the farad, F, which equals a coulomb per volt (C/V)

Capacitors and Charge Storage, continued
Capacitance Capacitors and Charge Storage, continued Capacitance is the ratio of charge to potential difference.

Capacitance Capacitors and Charge Storage, continued Capacitance depends on the size and shape of a capacitor. Capacitance for a Parallel-Plate Capacitor in a Vacuum

Capacitance Capacitors and Charge Storage, continued The material between a capacitor’s plates can change its capacitance. The effect of a dielectric is to reduce the strength of the electric field in a capacitor.

Capacitors in Keyboards
Capacitance Capacitors in Keyboards

Capacitance Energy and Capacitors The potential energy stored in a charged capacitor depends on the charge and the potential difference between the capacitor’s two plates.

Sample Problem Capacitance
A capacitor, connected to a 12 V battery, holds 36 µC of charge on each plate. What is the capacitance of the capacitor? How much electrical potential energy is stored in the capacitor? Given: Q = 36 µC = 3.6  10–5 C ∆V = 12 V Unknown: C = ? PEelectric = ?

Capacitance Sample Problem, continued Capacitance To determine the capacitance, use the definition of capacitance.

Capacitance Sample Problem, continued Capacitance To determine the potential energy, use the alternative form of the equation for the potential energy of a charged capacitor:

Current and Charge Movement
Current and Resistance Current and Charge Movement Electric current is the rate at which electric charges pass through a given area.

Current and Resistance
Drift Velocity Drift velocity is the the net velocity of a charge carrier moving in an electric field. Drift speeds are relatively small because of the many collisions that occur when an electron moves through a conductor.

Current and Resistance
Resistance to Current Resistance is the opposition presented to electric current by a material or device. The SI units for resistance is the ohm (Ω) and is equal to one volt per ampere. Resistance

Resistance to Current, continued
Current and Resistance Resistance to Current, continued For many materials resistance is constant over a range of potential differences. These materials obey Ohm’s Law and are called ohmic materials. Ohm’s low does not hold for all materials. Such materials are called non-ohmic. Resistance depends on length, cross-sectional area, temperature, and material.

Resistance to Current, continued
Current and Resistance Resistance to Current, continued Resistors can be used to control the amount of current in a conductor. Salt water and perspiration lower the body's resistance. Potentiometers have variable resistance.

Sources and Types of Current
Electric Power Sources and Types of Current Batteries and generators supply energy to charge carriers. Current can be direct or alternating. In direct current, charges move in a single direction. In alternating current, the direction of charge movement continually alternates.

Electric power = current  potential difference
Energy Transfer Electric power is the rate of conversion of electrical energy. Electric power P = I∆V Electric power = current  potential difference

Energy Transfer, continued
Electric Power Energy Transfer, continued Power dissipated by a resistor Electric companies measure energy consumed in kilowatt-hours. Electrical energy is transferred at high potential differences to minimize energy loss.

Multiple Choice 1. What changes would take place if the electron moved from point A to point B in the uniform electric field? A. The electron’s electrical potential energy would increase; its electric potential would increase. B. The electron’s electrical potential energy would increase; its electric potential would decrease. C. The electron’s electrical potential energy would decrease; its electric potential would decrease. D. Neither the electron’s electrical potential energy nor its electric potential would change.

1. What changes would take place if the electron moved from point A to point B in the uniform electric field? A. The electron’s electrical potential energy would increase; its electric potential would increase. B. The electron’s electrical potential energy would increase; its electric potential would decrease. C. The electron’s electrical potential energy would decrease; its electric potential would decrease. D. Neither the electron’s electrical potential energy nor its electric potential would change.

2. What changes would take place if the electron moved from point A to point C in the uniform electric field? F. The electron’s electrical potential energy would increase; its electric potential would increase. G. The electron’s electrical potential energy would increase; its electric potential would decrease. H. The electron’s electrical potential energy would decrease; its electric potential would decrease. J. Neither the electron’s electrical potential energy nor its electric potential would change.

Use the following passage to answer questions 3–4. A proton (q = 1.6  10–19 C) moves 2.0  10–6 m in the direction of an electric field that has a magnitude of 2.0 N/C. 3. What is the change in the electrical potential energy associated with the proton? A. –6.4  10–25 J B. –4.0  10–6 V C  10–25 J D  10–6 V

Use the following passage to answer questions 3–4. A proton (q = 1.6  10–19 C) moves 2.0  10–6 m in the direction of an electric field that has a magnitude of 2.0 N/C. 4. What is the potential difference between the proton’s starting point and ending point? F. –6.4  10–25 J G. –4.0  10–6 V H  10–25 J J  10–6 V

5. If the negative terminal of a 12 V battery is grounded, what is the potential of the positive terminal? A. –12 V B. +0 V C. +6 V D. +12 V

6. If the area of the plates of a parallel-plate capacitor is doubled while the spacing between the plates is halved, how is the capacitance affected? F. C is doubled G. C is increased by four times H. C is decreased by 1/4 J. C does not change

Use the following passage to answer questions 7–8. A potential difference of 10.0 V exists across the plates of a capacitor when the charge on each plate is 40.0 µC. 7. What is the capacitance of the capacitor? A  10–4 F B  10–4 F C  10–6 F D  10–6 F

Use the following passage to answer questions 7–8. A potential difference of 10.0 V exists across the plates of a capacitor when the charge on each plate is 40.0 µC. 8. How much electrical potential energy is stored in the capacitor? F  10–4 J G  10–4 J H  10–6 J J  10–6 J

9. How long does it take 5.0 C of charge to pass through a given cross section of a copper wire if I = 5.0 A? A s B. 1.0 s C. 5.0 s D. 25 s

10. A potential difference of 12 V produces a current of 0.40 A in a piece of copper wire. What is the resistance of the wire? F. 4.8 Ω G. 12 Ω H. 30 Ω J. 36 Ω

11. How many joules of energy are dissipated by a 50.0 W light bulb in 2.00 s? A J B J C. 100 J D. 200 J

12. How much power is needed to operate a radio that draws 7.0 A of current when a potential difference of 115 V is applied across it? F. 6.1  10–2 W G. 2.3  100 W H. 1.6  101 W J. 8.0  102 W

Short Response 13. Electrons are moving from left to right in a wire. No other charged particles are moving in the wire. In what direction is the conventional current?

13. Electrons are moving from left to right in a wire. No other charged particles are moving in the wire. In what direction is the conventional current? Answer: right to left

14. What is drift velocity, and how does it compare with the speed at which an electric field travels through a wire?

14. What is drift velocity, and how does it compare with the speed at which an electric field travels through a wire? Answer: Drift velocity is the net velocity of a charge carrier moving in an electric field. Drift velocities in a wire are typically much smaller than the speeds at which changes in the electric field propagate through the wire.

15. List four factors that can affect the resistance of a wire.

15. List four factors that can affect the resistance of a wire. Answer: length, cross-sectional area (thickness), temperature, and material

Extended Response 16. A parallel-plate capacitor is made of two circular plates, each of which has a diameter of 2.50  10–3 m. The plates of the capacitor are separated by a space of 1.40  10–4 m. a. Assuming that the capacitor is operating in a vacuum and that the permittivity of a vacuum (e0 = 8.85  10–12 C2/N•m2) can be used, determine the capacitance of the capacitor.

16. A parallel-plate capacitor is made of two circular plates, each of which has a diameter of 2.50  10–3 m. The plates of the capacitor are separated by a space of 1.40  10–4 m. a. Assuming that the capacitor is operating in a vacuum and that the permittivity of a vacuum (e0 = 8.85  10–12 C2/N•m2) can be used, determine the capacitance of the capacitor. Answer: 3.10  10–13 F

16. A parallel-plate capacitor is made of two circular plates, each of which has a diameter of 2.50  10–3 m. The plates of the capacitor are separated by a space of 1.40  10–4 m. b. How much charge will be stored on each plate of the capacitor when the capacitor’s plates are connected across a potential difference of 0.12 V?

16. A parallel-plate capacitor is made of two circular plates, each of which has a diameter of 2.50  10–3 m. The plates of the capacitor are separated by a space of 1.40  10–4 m. b. How much charge will be stored on each plate of the capacitor when the capacitor’s plates are connected across a potential difference of 0.12 V? Answer: 3.7  10–14 C

16. A parallel-plate capacitor is made of two circular plates, each of which has a diameter of 2.50  10–3 m. The plates of the capacitor are separated by a space of 1.40  10–4 m. c. What is the electrical potential energy stored in the capacitor when fully charged by the potential difference of 0.12 V?

16. A parallel-plate capacitor is made of two circular plates, each of which has a diameter of 2.50  10–3 m. The plates of the capacitor are separated by a space of 1.40  10–4 m. c. What is the electrical potential energy stored in the capacitor when fully charged by the potential difference of 0.12 V? Answer: 2.2  10–15 J

16. A parallel-plate capacitor is made of two circular plates, each of which has a diameter of 2.50  10–3 m. The plates of the capacitor are separated by a space of 1.40  10–4 m. d. What is the potential difference between a point midway between the plates and a point that is 1.10  10–4 m from one of the plates?

16. A parallel-plate capacitor is made of two circular plates, each of which has a diameter of 2.50  10–3 m. The plates of the capacitor are separated by a space of 1.40  10–4 m. d. What is the potential difference between a point midway between the plates and a point that is 1.10  10–4 m from one of the plates? Answer: 3.4  10–2 V

16. A parallel-plate capacitor is made of two circular plates, each of which has a diameter of 2.50  10–3 m. The plates of the capacitor are separated by a space of 1.40  10–4 m. e. If the potential difference of 0.12 V is removed from the circuit and the circuit is allowed to discharge until the charge on the plates has decreased to 70.7 percent of its fully charged value, what will the potential difference across the capacitor be?

16. A parallel-plate capacitor is made of two circular plates, each of which has a diameter of 2.50  10–3 m. The plates of the capacitor are separated by a space of 1.40  10–4 m. e. If the potential difference of 0.12 V is removed from the circuit and the circuit is allowed to discharge until the charge on the plates has decreased to 70.7 percent of its fully charged value, what will the potential difference across the capacitor be? Answer: 8.5  10–2 V

Capacitance Charging a Capacitor

A Capacitor With a Dielectric
Capacitance A Capacitor With a Dielectric

Schematic Diagrams and Circuits
A schematic diagram is a representation of a circuit that uses lines to represent wires and different symbols to represent components. Some symbols used in schematic diagrams are shown at right.

Electric Circuits An electric circuit is a set of electrical components connected such that they provide one or more complete paths for the movement of charges. A schematic diagram for a circuit is sometimes called a circuit diagram. Any element or group of elements in a circuit that dissipates energy is called a load.

Electric Circuits, continued
Schematic Diagrams and Circuits Electric Circuits, continued A circuit which contains a complete path for electrons to follow is called a closed circuit. Without a complete path, there is no charge flow and therefore no current. This situation is called an open circuit. A short circuit is a closed circuit that does not contain a load. Short circuits can be hazardous.

Schematic Diagrams and Circuits Electric Circuits, continued The source of potential difference and electrical energy is the circuits emf. Any device that transforms nonelectrical energy into electrical energy, such as a battery or a generator, is a source of emf. If the internal resistance of a battery is neglected, the emf equals the potential difference across the source’s two terminals.

Schematic Diagrams and Circuits Electric Circuits, continued The terminal voltage is the potential difference across a battery’s positive and negative terminals. For conventional current, the terminal voltage is less than the emf. The potential difference across a load equals the terminal voltage.

Light Bulb

Resistors in Series or in Parallel
A series circuit describes two or more components of a circuit that provide a single path for current. Resistors in series carry the same current. The equivalent resistance can be used to find the current in a circuit. The equivalent resistance in a series circuit is the sum of the circuit’s resistances. Req = R1 + R2 + R3…

Resistors in Series, continued
Resistors in Series or in Parallel Resistors in Series, continued Two or more resistors in the actual circuit have the same effect on the current as one equivalent resistor. The total current in a series circuit equals the potential difference divided by the equivalent resistance.

Sample Problem Resistors in Series
Resistors in Series or in Parallel Sample Problem Resistors in Series A 9.0 V battery is connected to four light bulbs, as shown at right. Find the equivalent resistance for the circuit and the current in the circuit.

Resistors in Series or in Parallel Sample Problem, continued Resistors in Series 1. Define Given: ∆V = 9.0 V R1 = 2.0 Ω R2 = 4.0 Ω R3 = 5.0 Ω R4 = 7.0 Ω Unknown: Req = ? I = ? Diagram:

Resistors in Series or in Parallel Sample Problem, continued Resistors in Series 2. Plan Choose an equation or situation: Because the resistors are connected end to end, they are in series. Thus, the equivalent resistance can be calculated with the equation for resistors in series. Req = R1 + R2 + R3… The following equation can be used to calculate the current. ∆V = IReq

Resistors in Series or in Parallel Sample Problem, continued Resistors in Series 2. Plan, continued Rearrange the equation to isolate the unknown: No rearrangement is necessary to calculate Req, but ∆V = IReq must be rearranged to calculate the current.

Resistors in Series or in Parallel Sample Problem, continued Resistors in Series 3. Calculate Substitute the values into the equation and solve: Substitute the equivalent resistance value into the equation for current.

Resistors in Series or in Parallel Sample Problem, continued Resistors in Series 4. Evaluate For resistors connected in series, the equivalent resistance should be greater than the largest resistance in the circuit. 18.0 Ω > 7.0 Ω

Resistors in Series, continued
Resistors in Series or in Parallel Resistors in Series, continued Series circuits require all elements to conduct electricity As seen below, a burned out filament in a string of bulbs has the same effect as an open switch. Because the circuit is no longer complete, there is no current.

Resistors in Parallel A parallel arrangement describes two or more com-ponents of a circuit that provide separate conducting paths for current because the components are connected across common points or junctions Lights wired in parallel have more than one path for current. Parallel circuits do not require all elements to conduct.

Resistors in Parallel

Resistors in Parallel, continued
Resistors in Series or in Parallel Resistors in Parallel, continued Resistors in parallel have the same potential differences across them. The sum of currents in parallel resistors equals the total current. The equivalent resistance of resistors in parallel can be calculated using a reciprocal relationship

Sample Problem Resistors in Parallel
Resistors in Series or in Parallel Sample Problem Resistors in Parallel A 9.0 V battery is connected to four resistors, as shown at right. Find the equivalent resistance for the circuit and the total current in the circuit.

Resistors in Series or in Parallel Sample Problem, continued Resistors in Parallel 1. Define Given: ∆V = 9.0 V R1 = 2.0 Ω R2 = 4.0 Ω R3 = 5.0 Ω R4 = 7.0 Ω Unknown: Req = ? I = ? Diagram:

Resistors in Series or in Parallel Sample Problem, continued Resistors in Parallel 2. Plan Choose an equation or situation: Because both sides of each resistor are connected to common points, they are in parallel. Thus, the equivalent resistance can be calculated with the equation for resistors in parallel. The following equation can be used to calculate the current. ∆V = IReq

Resistors in Series or in Parallel Sample Problem, continued Resistors in Parallel 2. Plan, continued Rearrange the equation to isolate the unknown: No rearrangement is necessary to calculate Req; rearrange ∆V = IReq to calculate the total current delivered by the battery.

Resistors in Series or in Parallel Sample Problem, continued Resistors in Parallel 3. Calculate Substitute the values into the equation and solve:

Resistors in Series or in Parallel Sample Problem, continued Resistors in Parallel 3. Calculate, continued Substitute the equivalent resistance value into the equation for current.

Resistors in Series or in Parallel Sample Problem, continued Resistors in Parallel 4. Evaluate For resistors connected in parallel, the equivalent resistance should be less than the smallest resistance in the circuit. 0.917 Ω < 2.0 Ω

Resistors Combined Both in Parallel and in Series
Complex Resistor Combinations Resistors Combined Both in Parallel and in Series Many complex circuits can be understood by isolating segments that are in series or in parallel and simplifying them to their equivalent resistances. Work backward to find the current in and potential difference across a part of a circuit.

Sample Problem Equivalent Resistance
Complex Resistor Combinations Sample Problem Equivalent Resistance Determine the equivalent resistance of the complex circuit shown below.

Complex Resistor Combinations Sample Problem, continued Equivalent Resistance Reasoning The best approach is to divide the circuit into groups of series and parallel resistors. This way, the methods presented in Sample Problems A and B can be used to calculate the equivalent resistance for each group.

Complex Resistor Combinations Sample Problem, continued Equivalent Resistance 1. Redraw the circuit as a group of resistors along one side of the circuit. Because bends in a wire do not affect the circuit, they do not need to be represented in a schematic diagram. Redraw the circuit without the corners, keeping the arrangement of the circuit elements the same. TIP: For now, disregard the emf source, and work only with the resistances.

Complex Resistor Combinations Sample Problem, continued Equivalent Resistance 2. Identify components in series, and calcu-late their equivalent resistance. Resistors in group (a) and (b) are in series. For group (a): Req = 3.0 Ω Ω = 9.0 Ω For group (b): Req = 6.0 Ω Ω = 8.0 Ω

Complex Resistor Combinations Sample Problem, continued Equivalent Resistance 3. Identify components in parallel, and calculate their equivalent resis-tance. Resistors in group (c) are in parallel.

Complex Resistor Combinations Sample Problem, continued Equivalent Resistance 4. Repeat steps 2 and 3 until the resistors in the circuit are reduced to a single equivalent resistance.The remainder of the resistors, group (d), are in series.

Sample Problem Current in and Potential Difference Across a Resistor
Complex Resistor Combinations Sample Problem Current in and Potential Difference Across a Resistor Determine the current in and potential difference across the 2.0 Ω resistor highlighted in the figure below.

Complex Resistor Combinations Sample Problem, continued Current in and Potential Difference Across a Resistor Reasoning First determine the total circuit current by reducing the resistors to a single equivalent resistance. Then rebuild the circuit in steps, calculating the current and potential difference for the equivalent resistance of each group until the current in and potential difference across the 2.0 Ω resistor are known.

Complex Resistor Combinations Sample Problem, continued Current in and Potential Difference Across a Resistor 1. Determine the equivalent resistance of the circuit. The equivalent resistance of the circuit is 12.7 Ω, as calculated in the previous Sample Problem.

Complex Resistor Combinations Sample Problem, continued Current in and Potential Difference Across a Resistor 2. Calculate the total current in the circuit. Substitute the potential difference and equivalent resistance in ∆V = IR, and rearrange the equation to find the current delivered by the battery.

Complex Resistor Combinations Sample Problem, continued 3. Determine a path from the equivalent resistance found in step 1 to the 2.0 Ω resistor. Review the path taken to find the equivalent resistance in the figure at right, and work backward through this path. The equivalent resistance for the entire circuit is the same as the equivalent resistance for group (d). The center resistor in group (d) in turn is the equivalent resistance for group (c). The top resistor in group (c) is the equivalent resistance for group (b), and the right resistor in group (b) is the 2.0 Ω resistor.

Complex Resistor Combinations Sample Problem, continued Current in and Potential Difference Across a Resistor 4. Follow the path determined in step 3, and calculate the current in and potential difference across each equivalent resistance. Repeat this process until the desired values are found.

Complex Resistor Combinations Sample Problem, continued 4. A. Regroup, evaluate, and calculate. Replace the circuit’s equivalent resistance with group (d). The resistors in group (d) are in series; therefore, the current in each resistor is the same as the current in the equivalent resistance, which equals 0.71 A. The potential difference across the 2.7 Ω resistor in group (d) can be calculated using ∆V = IR. Given: I = 0.71 A R = 2.7 Ω Unknown: ∆V = ? ∆V = IR = (0.71 A)(2.7 Ω) = 1.9 V

Complex Resistor Combinations Sample Problem, continued 4. B. Regroup, evaluate, and calculate. Replace the center resistor with group (c). The resistors in group (c) are in parallel; therefore, the potential difference across each resistor is the same as the potential difference across the 2.7 Ω equivalent resistance, which equals 1.9 V. The current in the 8.0 Ω resistor in group (c) can be calculated using ∆V = IR. Given: ∆V = 1.9 V R = 8.0 Ω Unknown: I = ?

Complex Resistor Combinations Sample Problem, continued 4. C. Regroup, evaluate, and calculate. Replace the 8.0 Ω resistor with group (b). The resistors in group (b) are in series; therefore, the current in each resistor is the same as the current in the 8.0 Ω equivalent resistance, which equals 0.24 A. The potential difference across the 2.0 Ω resistor can be calculated using ∆V = IR. Given: I = 0.24 A R = 2.0 Ω Unknown: ∆V = ?

Multiple Choice 1. Which of the following is the correct term for a circuit that does not have a closed-loop path for electron flow? A. closed circuit B. dead circuit C. open circuit D. short circuit

1. Which of the following is the correct term for a circuit that does not have a closed-loop path for electron flow? A. closed circuit B. dead circuit C. open circuit D. short circuit

2. Which of the following is the correct term for a circuit in which the load has been unintentionally bypassed? F. closed circuit G. dead circuit H. open circuit J. short circuit

Use the diagram below to answer questions 3–5. 3. Which of the circuit elements contribute to the load of the circuit? A. Only A B. A and B, but not C C. Only C D. A, B, and C

Use the diagram below to answer questions 3–5. 4. Which of the following is the correct equation for the equivalent resis-tance of the circuit?

Use the diagram below to answer questions 3–5. 5. Which of the following is the correct equation for the current in the resistor?

Use the diagram below to answer questions 6–7. 6. Which of the following is the correct equation for the equivalent resis-tance of the circuit?

Use the diagram below to answer questions 6–7. 7. Which of the following is the correct equation for the current in resistor B?

8. Three 2.0 Ω resistors are connected in series to a 12 V battery. What is the potential difference across each resistor? F. 2.0 V G. 4.0 V H. 12 V J. 36 V

Use the following passage to answer questions 9–11. Six light bulbs are connected in parallel to a 9.0 V battery. Each bulb has a resistance of 3.0 Ω. 9. What is the potential difference across each bulb? A. 1.5 V B. 3.0 V C. 9.0 V D. 27 V

Use the following passage to answer questions 9–11. Six light bulbs are connected in parallel to a 9.0 V battery. Each bulb has a resistance of 3.0 Ω. 10. What is the current in each bulb? F. 0.5 A G. 3.0 A H. 4.5 A J. 18 A

Use the following passage to answer questions 9–11. Six light bulbs are connected in parallel to a 9.0 V battery. Each bulb has a resistance of 3.0 Ω. 11. What is the total current in the circuit? A. 0.5 A B. 3.0 A C. 4.5 A D. 18 A

Short Response 12. Which is greater, a battery’s terminal voltage or the same battery’s emf? Explain why these two quantities are not equal.

12. Which is greater, a battery’s terminal voltage or the same battery’s emf? Explain why these two quantities are not equal. Answer: A battery’s emf is slightly greater than its terminal voltage. The difference is due to the battery’s internal resistance.

13. Describe how a short circuit could lead to a fire.

13. Describe how a short circuit could lead to a fire. Answer: In a short circuit, the equivalent resistance of the circuit drops very low, causing the current to be very high. The higher current can cause wires still in the circuit to overheat, which may in turn cause a fire in materials contacting the wires.

14. Explain the advantage of wiring the bulbs in a string of decorative lights in parallel rather than in series.

14. Explain the advantage of wiring the bulbs in a string of decorative lights in parallel rather than in series. Answer: If one bulb is removed, the other bulbs will still carry current.

Extended Response 15. Using standard symbols for circuit elements, draw a diagram of a circuit that contains a battery, an open switch, and a light bulb in parallel with a resistor. Add an arrow to indicate the direction of current if the switch were closed.

15. Using standard symbols for circuit elements, draw a diagram of a circuit that contains a battery, an open switch, and a light bulb in parallel with a resistor. Add an arrow to indicate the direction of current if the switch were closed. Answer:

Use the diagram below to answer questions 16–17. 16. For the circuit shown, calculate the following: a. the equivalent resistance of the circuit b. the current in the light bulb. Show all your work for both calculations.

Use the diagram below to answer questions 16–17. 16. For the circuit shown, calculate the following: a. the equivalent resistance of the circuit b. the current in the light bulb. Show all your work for both calculations. Answer: a. 4.2 Ω b. 2.9 A

Use the diagram below to answer questions 16–17. 17. After a period of time, the 6.0 Ω resistor fails and breaks. Describe what happens to the brightness of the bulb. Support your answer.

Use the diagram below to answer questions 16–17. 17. Answer: The bulb will grow dim. The loss of the 6.0 Ω resistor causes the equivalent resistance of the circuit to increase to 4.5 Ω. As a result, the current in the bulb drops to 2.7 A, and the brightness of the bulb decreases.

18. Find the current in and potential difference across each of the resistors in the following circuits: a. a 4.0 Ω and a 12.0 Ω resistor wired in series with a 4.0 V source. b. a 4.0 Ω and a 12.0 Ω resistor wired in parallel with a 4.0 V source. Show all your work for each calculation.

18. Find the current in and potential difference across each of the resistors in the following circuits: a. a 4.0 Ω and a 12.0 Ω resistor wired in series with a 4.0 V source. b. a 4.0 Ω and a 12.0 Ω resistor wired in parallel with a 4.0 V source. Show all your work for each calculation. Answers: a. 4.0 Ω: 0.25 A, 1.0 V 12.0 Ω: 0.25 A, 3.0 V b. 4.0 Ω: 1.0 A, 4.0 V 12.0 Ω: 0.33 A, 4.0 V

19. Find the current in and potential difference across each of the resistors in the following circuits: a. a 150 Ω and a 180 Ω resistor wired in series with a 12 V source. b. a 150 Ω and a 180 Ω resistor wired in parallel with a 12 V source. Show all your work for each calculation.

19. Find the current in and potential difference across each of the resistors in the following circuits: a. a 150 Ω and a 180 Ω resistor wired in series with a 12 V source. b. a 150 Ω and a 180 Ω resistor wired in parallel with a 12 V source. Show all your work for each calculation. Answer: a.150 Ω: A, 5.4 V 180 Ω: A, 6.5 V b. 150 Ω: A, 12 V 180 Ω: A, 12 V

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Properties of Electric Charge

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