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Ch. 20 Electric Potential and Electric Potential Energy

Ch. 20 Electric Potential and Electric Potential Energy. Electric Potential Energy.

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Ch. 20 Electric Potential and Electric Potential Energy

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  1. Ch. 20 Electric Potential and Electric Potential Energy

  2. Electric Potential Energy • Electrical potential energy is the energy contained in a configuration of charges. Like all potential energies, when it goes up the configuration is less stable; when it goes down, the configuration is more stable. • The unit is the Joule.

  3. Equation • DU = - W = q0Ed • q0 – test charge • E – Electric field • d - distance

  4. Electrical potential energy increaseswhen charges are brought into lessfavorable configurations

  5. Electrical potential energy decreases when charges are brought into more favorable configurations.

  6. Work must be done on the charge to increase the electric potential energy

  7. For a positive test charge to be moved upward a distance d, the electric force does negative work. • The electric potential energy has increased and U is positive (U2 > U1)

  8. If a negative charge is moved upward a distance d, the electric force does positive work. • The change in the electric potential energy U is negative (U2 < U1)

  9. Electric Potential (V) Electric potential is hard to understand, but easy to measure. • We commonly call it “voltage”, and its unit is the Volt. • 1 V = 1 J/C • Electric potential is easily related to both the electric potential energy, and to the electric field.

  10. The change in potential energy is directly related to the change in voltage. DU = qDV DV = DU/q • DU: change in electrical potential energy (J) • q: charge moved (C) • DV: potential difference (V) • All charges will spontaneously go to lower potential energies if they are allowed to move.

  11. Since all charges try to decrease UE, and DUE = qDV, this means that spontaneous movement of charges result in negative DU. • DV = DU / q • Positive charges like to DECREASE their potential (DV < 0) • Negative charges like to INCREASE their potential. (DV > 0)

  12. Sample Problem: A 3.0 μC charge is moved through a potential difference of 640 V. What is its potential energy change?

  13. Sample Problem: A 3.0 μC charge is moved through a potential difference of 640 V. What is its potential energy change?

  14. Electrical Potential in Uniform Electric Fields The electric potential is related in a simple way to a uniform electric field. DV = -Ed • DV: change in electrical potential (V) • E: Constant electric field strength (N/C or V/m) • d: distance moved (m)

  15. Sample Problem: An electric field is parallel to the x-axis. What is its magnitude and direction if the potential difference between x =1.0 m and x = 2.5 m is found to be +900 V?

  16. Sample Problem: An electric field is parallel to the x-axis. What is its magnitude and direction if the potential difference between x =1.0 m and x = 2.5 m is found to be +900 V?

  17. Sample Problem: If a proton is accelerated through a potential difference of 2.000 V, what is its change in potential energy? How fast will this proton be moving if it started at rest?

  18. Sample Problem: A proton at rest is released in a uniform electric field. What potential difference must it move through in order to acquire a speed of 0.20 c?

  19. Electric Potential Energy forSpherical Charges • Electric potential energy is a scalar, like all forms of energy. U = kq1q2/r • U: electrical potential energy (J) • k: 8.99 × 109 N m2 / C2 • q1, q2 : charges (C) • r: distance between centers (m) This formula only works for spherical charges or point charges.

  20. Absolute Electric Potential(spherical) • For a spherical or point charge, the electric potential can be calculated by the following Formula: V = kq/r • V: potential (V) • k: 8.99 x 109 N m2/C2 • q: charge (C) • r: distance from the charge (m) • Remember, k = 1/(4peo)

  21. Electric Field and Electric Potential E = - V / d Two things about E and V: • The electric field points in the direction of decreasing electric potential. • The electric field is always perpendicular to the equipotential surface.

  22. 20-4 Equipotential Surfaces and the Electric Field For two point charges:

  23. Equipotential Surfaces and the Electric Field An ideal conductor is an equipotential surface. Therefore, if two conductors are at the same potential, the one that is more curved will have a larger electric field around it. This is also true for different parts of the same conductor.

  24. Equipotential Surfaces and the Electric Field There are electric fields inside the human body; the body is not a perfect conductor, so there are also potential differences. An electrocardiograph plots the heart’s electrical activity.

  25. Equipotential Surfaces and the Electric Field An electroencephalograph measures the electrical activity of the brain:

  26. Capacitor • Named for the capacity to store electric charge and energy. • A capacitor is two conducting plates separated by a finite distance:

  27. The capacitance relates the charge to the potential difference:

  28. Sample Problem: A 0.75 F capacitor is charged to a voltage of 16 volts. What is the magnitude of the charge on each plate of the capacitor?

  29. Sample Problem: A 0.75 mF capacitor is charged to a voltage of 16 volts. What is the magnitude of the charge on each plate of the capacitor? V = 16 V, C = 0.75 mF = 0.75 x 10-6 F Q =? C = Q/V or Q = CV Q = (0.75 x 10-6)(16) Q = 1.2 x 10-5 C

  30. A simple type of capacitor is the parallel-plate capacitor. It consists of two plates of area A separated by a distance d. By calculating the electric field created by the charges ±Q, we find that the capacitance of a parallel-plate capacitor is:

  31. The general properties of a parallel-plate capacitor – that the capacitance increases as the plates become larger and decreases as the separation increases – are common to all capacitors.

  32. Capacitor Geometry The capacitance of a capacitor depends on HOW you make it.

  33. Sample Problem: What is the AREA of a 1 F capacitor that has a plate separation of 1 mm? C = 1 F, d = 1 mm = 0.001 m, eo= 8.85 x 10-12 C2/(Nm2)

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