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Chapter 21 Electric Fields

21.1 Creating and Measuring Electric Fields. Coulomb's LawThe magnitude of the electric force (F) between two charges q1 and q2, separated by a distance r, is proportional to the magnitude of the charges and inversely proportional to the square of the distance. . . Where k is a constant. the SI unit for charge is the coulomb (C). When r is measured in meters, and F is measured in N, then k = 9

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Chapter 21 Electric Fields

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    2. 21.1 Creating and Measuring Electric Fields Coulomb’s Law The magnitude of the electric force (F) between two charges q1 and q2, separated by a distance r, is proportional to the magnitude of the charges and inversely proportional to the square of the distance.

    3. Example Problem Two charges are separated by 3.0 cm. Object A has a charge of + 6 ?C, while object B has a charge of + 3.0 ?C. What is the force on object A?(180 N)

    4. Electric Field (symbol: E): is produced by a charged particle and is invisible. It is a vector quantity that relates the force exerted on a test charge to the size of the test charge. A charge “source charge” creates an electric field about it in all directions. We can detect the electric field presence by observing that a charged particle (test charge) placed in an electric field will always experience a force which will tend to “push” or “pull” that charged particle in some particular direction. We assume that the test charge is so small that it has no influence on the source charges, and that it doesn’t alter the existing electric field by moving the source charge to a new location.

    5. Magnitude and direction of the electric field: The magnitude and direction of the electric field at a particular point in space can be found by placing a test charge at this point. The amount of charge on the test charge is symbolized by qtest. The magnitude of the electric field at a point A is given by this equation: The direction of the electric field at point A is defined to be the same as the direction of the force experienced by a positive test charge placed at that point.

    6. Example Problem: An electric field is to be measured using a positive test charge of 4.0 × 10-5 C. This test charge experiences a force of 0.60 N acting at an angle of 10°. What is the magnitude and direction of the electric field at the location of the test charge?

    9. 21.2 Applications of Electric Fields A particle with a charge q placed in an electric field will be subject to electrostatic forces and will have a potential energy. Moving the charge will change its potential energy from PEa to PEb, reflecting the work Wba done by the electric field. Potential energy depends also on the magnitude of the test charge being transported. The larger the test charge, the greater the increase in its potential energy

    10. The difference Vab between the potentials at two points a and b is the voltage. Voltage provides a measure of the work per unit charge required to move the charge between two points a and b in the field; it represents the corresponding difference in potential energy per unit charge. ?Vab = Va – Vb = Wba/q = PEa /q- PEb /q Electric potential and Voltage are measured in units of volt (V) or (J/C). For a current-carrying wire, the potential difference between two points along the wire causes the current to flow in that segment.

    11. Batteries, for example, are rated by the potential difference across their terminals. In a nine volt battery the potential difference between the positive and negative terminals is precisely nine volts. On the other hand the potential difference across the power outlet in the wall of your home is 110 volts. Electric potential differences are measured with a voltmeter.

    12. The Electric Potential in a Uniform Field A Uniform electric field can be made by placing two large, flat conducting plates, oppositely charged, parallel to each other. The electric field direction is from positive to the negative plate. If a positive charge qtest is moved a distance d, in the direction opposite the electric field direction, the work done is: W on qtest = Fd, ?V=Fd/ qtest = (F/ qtest)d = Ed

    13. Example problem#1 Two parallel plates are given opposite charges. A voltmeter measures the electric potential difference to be 60.0 V. The plates are 3.0 cm apart. What is the magnitude of the electric field between them?

    14. Example Problem # 2 Two large charged parallel plates are 4.0 cm apart. The magnitude of the electric field between the plates is 625 N/C. What is the electric potential difference between the plates? What work will you do to move a charge equal to that of one proton (q=1.6 × 10-19 C) from the negative to the positive plate?

    15. Storing Charges: The Capacitor

    17. Grounding The potential of the earth is zero. Any object connected to the earth will have its excess charge flow into the earth. It is considered to be grounded.

    18. Capacitor A device (sometimes called a condenser) that stores charge in the electric field between its plates. Each plate carries the same amount of charge, one plate being negative and the other being positive. A potential difference exists between the two plates. Capacitor consists of two parallel metals plates insulated from each other by a dielectric, a material that does not conduct electricity. Capacitors are used in electric circuits to store charges.

    19. Capacitance Is the ability of capacitor to store electric charge symbol is C and SI unit is the Farad, F C = q/?V where q is the charge stored in Coulombs (+q on one plate and –q on the other), C is the capacitance, and ?V is the potential difference between the conducting surfaces in volt. Capacitance depends only on the construction of the capacitor, not on the charge, q.

    20. Example Problem A sphere has an electric potential difference between it and Earth of 60.0 V when it has been charged to 3.0 × 10-6 C. What is its capacitance?

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