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Electric Energy and Circuits Electrostatic Equilibrium • No net motion of charge is occurring within a conductor • Meets the following conditions ▫ Electric field is zero everywhere inside the conductor ▫ Any excess charge on an isolated conductor resides entirely on on the conductor’s outer surface Electrostatic Equilibrium ▫ 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 (i.e. Sharp points.) ▫ Electrostatic Equilibrium = “Equipotential” Electric Potential Energy, U • Electric potential energy –the change in electrical potential energy of a charge ▫ U = Uf - Ui = -W ▫ The work done by a conservative force (e.g., an electric force and force of gravity) is equal to the negative of the change in potential energy. ▫ SI units: Joules (J) • Similar to energy due to an objects position above the earth (GPE) ▫ Results from the interaction of two objects’ charges; not mass = − Ed ∆U = -W therefore: ∆U = − Ed (a) A positive test charge q0, experiences a downward force due to the electric field . If the charge is moved upward a distance d, the work done by the electric field is –qoEd. At the same time, the electric potential energy of the system increases by qoEd. The situation is analogous to that of an object in gravitational field. (b) If a ball is lifted against the force exerted by gravity, the gravitational potential energy of the system increases. Electric Potential Energy ………final thoughts • Work is done any time charge moves b/c of an electric force W = -∆U (U = potential energy) • Electric potential energy is conserved SI = joule (J) • q = charge • E = strength of electric field • d = displacement ▫ Magnitude of displacement’s component in the direction of the eclectic field. Electric Potential Energy • (-) indicates that PEelectric will↑ if (-q) and ↓ if (+q) • Valid only in a uniform electric field • Any displacement perpendiular to and electric field does not change the PEelectric Peelectric In a Uniform Electric Field Toward E Opposite E + charge - charge Loses Peelectric Gains Peelectric Gains Peelectric Loses PEelectric Electric Potential, V • Electric Potential: the change in electric potential: ▫ V = Vf – Vi = U/q0 = (-W)/q0 ▫ V is the change in electric potential energy per charge. ▫ SI units: Joules/Coulomb (J/C) = Volt (V) • Relationship between U and V: ▫ U = q0 V ▫ Both are scalar quantities. Electric Potential, V • Another commonly used unit of energy is the electron volt (eV): • 1 eV = (1.60x10-19 C)(1 V) = 1.60x10-19 J Practice Problem Exercise 20-1 pg. 664 AP book Find the change in electric potential energy, U, as a charge of (a) 2.20 x 10-6 C or (b) -1.10 x 10-6 C moves from a point A to a point B, given that the change in electric potential between these points is V = VB – VA = 24.0 V. Connection between Electric Field and Electric Potential • Connection between the electric field and the electric potential: ▫ E = -V/s ▫ V = -Es • The electric field depends on : rate of change of the electric potential with position. • The electric potential decreases as one moves in the direction of the electric field. • SI units for E: 1 N/C = 1 V/m Practice Problem: Plates at Different Potentials Example 20-1 AP book pg. 665 A uniform electric field is established by connecting the plates of a parallel-plate capacitor to a 12-V battery. (a) If d = 0.75 cm, what is the magnitude of the electric field in the capacitor? (b) A charge of +6.24x10-6 C moves from the positive plate to the negative plate. Find the change in electric potential energy. (In electrical systems, we shall assume that gravity can be ignored, unless specifically instructed otherwise.) Energy Conservation Energy conservation: For a charged object in an electric field, its total energy must be conserved. KA + UA = KB + UB, or (1/2)mvA2 + UA = (1/2)mvB2 + UB Practice Problem: From Plate to Plate Example 20-2 AB book pg. 668 • Suppose a charge q = +6.24 x 10-6 C is released from rest at the positive plate and that it reaches the negative plate with a speed of 3.4 m/s. • (a) What is the mass of the charge? • (b) What is its final kinetic energy? The Electric Potential of Point Charges Exercise 20-2 AP book pg. 670 Electric potential V produced by a point charge q at a distance r: Conventionally, choosing the electric potential to be zero at infinity, V = kq/r Electric potential energy U for point charges q and q0 separated by a distance r: U = q0V = kq0q/r The Electric Potential of Point Charges Exercise 20-2 AP book pg. 670 Find the electric potential by a point charge of 6.80x10-7 C at a distance of 2.60 m. “U’ associated with multiple charges • If a second charge is placed nearby, there will be Peelectric associated with the two charges. • Electric field is not uniform “U” associated with multiple charges • Point of reference for Peelectric is assumed to be infinity ▫ Peelectric goes to zero as r btwn q’s goes to infinity • Like charges repel – (+) W must be done to bring them together ▫ Peelectric is (+) for like charges and (-) for opposite charges 20 Electric Fields in Circuits • Point away from positive terminal, towards negative • Channeled by conductor (wire) • Electrons flow opposite field lines (neg. charge) electrons & direction of motion E E E E Electric field direction