Section 2: Applications of Electric Fields Electric potential (sometimes called voltage) is electric potential energy per unit charge. K What I Know W What I Want to Find Out L What I Learned
Essential Questions What is an electric potential difference? How is potential difference related to the work required to move a charge? What are properties of capacitors? Copyright © McGraw-Hill Education Applications of Electric Fields
Vocabulary Review New work electric potential difference volt equipotential capacitor capacitance Applications of Electric Fields Copyright © McGraw-Hill Education
Energy and Electric Potential Concepts in Motion Copyright © McGraw-Hill Education Applications of Electric Fields
Energy and Electric Potential The electric potential difference (ΔV), which often is called potential difference, is the work (Won q’) needed to move a positive test charge from one point to another, divided by the magnitude of the test charge. You also can think of electric potential difference as the change in electric potential energy (ΔPE) per unit charge. ΔV is measured in joules per coulomb (J/C). One (J/C) is called a volt (V). Whenever the electric potential difference between two or more positions is zero, those positions are said to be at equipotential. Electric Potential Difference Copyright © McGraw-Hill Education Applications of Electric Fields
Electric Potential in a Uniform Field You can produce a uniform electric field by placing two large, flat conducting plates parallel to each other. One plate is charged positively, and the other plate is charged negatively. The magnitude and the direction of the electric field are the same at all points between the plates, except at the edges of the plates, and the electric field points from the positive plate to the negative plate. The electric potential is higher near the positively charged plate and lower near the negatively charged plate. You can represent the electric potential difference (ΔV) between two points a distance (d) apart in a uniform field (E) with the following equation. Electric Potential Difference in a Uniform Field Copyright © McGraw-Hill Education Applications of Electric Fields
Electric Potential in a Uniform Field Use with Example Problem 3. Problem Two 30-cm-square charged parallel plates are 4.0 cm apart. The magnitude of the electric field between the plates is 2400 N/C. What is the electric potential difference between the two fields? How much work is required to move a proton from the negative plate to the positive plate? SOLVE FOR THE UNKNOWN Use the relationship among the electric field, the plate separation, and the potential difference for a uniform field. Use the relationship among the work, charge, and potential difference. Response SKETCH AND ANALYZE THE PROBLEM Sketch the situation. List the knowns and unknowns. KNOWN UNKNOWN E = 2400 N/C ΔV = ? d = 4.0 cm W = ? q = 1.602×10−19 C EVALUATE THE ANSWER The answers have the correct units. Potential difference is in volts and work is in joules. Applications of Electric Fields Copyright © McGraw-Hill Education
Millikan’s Oil-Drop Experiment Concepts in Motion Copyright © McGraw-Hill Education Applications of Electric Fields
Millikan’s Oil-Drop Experiment Use with Example Problem 4. Problem An oil drop suspended motionless between two 30-cm-square parallel plates weighs 1.510−14 N. The parallel plates are 2.4 cm apart, and the potential difference between them is 450 V. What is the charge on the oil drop? If the upper plate is positive, how many excess electrons are on the oil drop? SOLVE FOR THE UNKNOWN The net force on the drop is zero, so the electric force and the gravitational force are equal in magnitude. Use this to find the charge on the drop. Response SKETCH AND ANALYZE THE PROBLEM Sketch the situation. List the knowns and unknowns. KNOWN UNKNOWN ΔV = 450 V Fg = 1.5×10−14 N q = ? d = 0.024 m e = 1.602×10−19 C n = ? Applications of Electric Fields Copyright © McGraw-Hill Education
Millikan’s Oil-Drop Experiment Use with Example Problem 4. Problem An oil drop suspended motionless between two 30-cm-square parallel plates weighs 1.510−14 N. The parallel plates are 2.4 cm apart, and the potential difference between them is 450 V. What is the charge on the oil drop? If the upper plate is positive, how many excess electrons are on the oil drop? SOLVE FOR THE UNKNOWN The charge on the drop is q = 8.0×10−19 C. Find the number of excess electrons. Response SKETCH AND ANALYZE THE PROBLEM Sketch the situation. List the knowns and unknowns. EVALUATE THE ANSWER Charge is measured in coulombs, so the units are correct. 5 excess electrons is reasonable for an oil drop in this setup. KNOWN UNKNOWN ΔV = 450 V Fg = 1.5×10−14 N q = ? d = 0.024 m e = 1.602×10−19 C n = ? Copyright © McGraw-Hill Education Applications of Electric Fields
Electric Fields Near Conductors Because electrons have like charges, they repel each other. In a conductor they are free to move, so they spread far apart in a way that minimizes their potential energy. The charges come to rest on the surface of the conductor. It does not matter if the conducting sphere is solid or hollow. The electric field is zero everywhere inside a closed, charged metal container. The electric field at the surface depends on the shape of the conductor; free charges are closer together at the sharp points of a conductor. Applications of Electric Fields Copyright © McGraw-Hill Education
Capacitors Capacitance Energy can be stored in an electric field. A device for storing electrical energy is called a capacitor. If you connect a 1-V power supply across a capacitor, the potential difference between the two plates would be 1 V. This would result in a net positive charge (+q) on one plate and a net negative charge of equal magnitude (-q) on the other plate. The graph of q v. ΔV is a straight line. The slope of the line in a net charge versus potential difference graph is a constant and is called the capacitance (C) of the capacitor. Capacitance is measured in farads (F), where 1 F = 1 C/V. Capacitance Copyright © McGraw-Hill Education Applications of Electric Fields
Capacitors Problem Response ΔV = 76.0 V q = 3.8×10−4 C Use with Example Problem 5. Problem The electric potential difference between a charged sphere and Earth is 76.0 V when the sphere has been charged to 3.8×10−4 C. What is the capacitance of the sphere-Earth system? SOLVE FOR THE UNKNOWN Use the relationship among capacitance, charge, and potential difference. Response SKETCH AND ANALYZE THE PROBLEM Sketch the situation. List the knowns and unknowns. KNOWN UNKNOWN q = 3.8×10−4 C C = ? ΔV = 76.0 V EVALUATE THE ANSWER Capacitance is in farads, so the units are correct. 5 microfarads is comparable to the capacitance of commercial capacitors. Copyright © McGraw-Hill Education Applications of Electric Fields
Capacitors Capacitors have many shapes and sizes and are used in computers, televisions, and digital cameras. The capacitance of a capacitor can be controlled by varying the surface area of the two conductors, or plates, by varying the distance between the plates, and by the nature of the insulating material between the plates, such as ceramic, mica, polyester, and paper. Just as 1 C is a large amount of net charge, 1 F is also a fairly large capacitance. Most capacitors used in modern electronics have capacitances between 10 picofarads (10×10−12 F) and 500 microfarads (500×10−6 F). However, the memory capacitors that are used to prevent loss of memory in some computers can have capacitance from 0.5 F to 1.0 F. Applications of Electric Fields Copyright © McGraw-Hill Education
Review Essential Questions Vocabulary What is an electric potential difference? How is potential difference related to the work required to move a charge? What are properties of capacitors? Vocabulary electric potential difference volt equipotential capacitor capacitance Copyright © McGraw-Hill Education Applications of Electric Fields