Physics 222 Exam Review Outline • Overview/Mind-map • What each equation does • Practice Problems Sorry about the boring theme. • I couldn’t find a suitable theme that I liked, that didn’t mess up my text. • However, looking at green is said to increase creativity and stimulate brain function. May use this later. Useful tip: Storing variables in calculator • ππ = 9 ∗ 109 • π0 = 8.85 ∗ 10−12 π¬ = −πV Electric Force Multiply by q Electric Potential ππ − ππ Potential Energy π΄ =− Divide by q Electric Field π¬ ⋅ ππ π΅ Capacitors Interacts with dipole Dipole Resistors Charge Overview • • • • • • Fluids Electric force -> Electric field Potential energy -> Electric potential E->V and V->E Capacitors and energy stored inside them Resistors πΉ π= π΄ • This is the definition of Pressure: Force/Area π = π0 + πππ¦ • Used to calculate the pressure at a depth of y in some medium, usually water. • Example: 5 m deep under the water, pressure=1.01 ∗ 105 + 103 ∗ 9.8 ∗ 5 ππ = π΄π£ = ππππ π‘πππ‘ ππ‘ • This is the continuity equation for fluid flow. • In English, it means that the amount of stuff going through a pipe is constant, so shrinking the pipe means that the water will go faster. 1 2 π + πππ¦ + ππ£ = ππππ π‘πππ‘ 2 • Also known as Bernoulli’s Equation • Classic problem: calculating velocity of water shooting out of a hole in a container. – In the case that the radius of the hole is small, the velocity at the bottom simplifies to π£ = 2ππ¦ , where y is the distance from the water level to the hole. Questions? ππΆππ’ππππ π1 π2 = ππ 2 π π π π¬= π0 • Just like electric force, but without the test charge q0. It’s still a vector. π π¬ = ππ 2 π π • Electric field of a point charge. Electric field lines go from + to -. • Also, line density indicates field strength π = |π|π • Dipole moment, BY DEFINITION • Notice: Dipole moment points from negative to positive….opposite of the direction E points π=π×π¬ • Torque on a dipole by the external electric field – Note that E is not the E produced by the dipole – E is external • Torque is maximum when dipole moment and E are perpendicular. π = −π ⋅ π¬ • Potential energy is minimum (also called stable equilibrium) when two things are true: – Dipole moment is parallel to E – Dipole moment points in the same direction as E. • Potential energy is maximum (also called unstable equilibrium) when two things are true: – Dipole moment is parallel to E. – Dipole moment points in the opposite direction as E. Questions? ππππ ΦπΈ = π¬ ⋅ ππ = π0 π π’πππππ • Also known as Gauss’s Law • Really there are two equations here…but they’re both equally valid…always. Steps in solving Gauss’s Law Problems – Draw a picture of the object. Pick a good Gaussian surface. – Write down the expression of Gauss’s Law that involves the dot product between E and A. (If E is perpendicular to A, the flux is 0 for that surface. Otherwise, use symmetry to get rid of the integral.) – Write down the expression of Gauss’s Law that involves the total charge q. – Set the two expressions equal to each other and eliminate variables. 2π πΈ = ππ π π • Electric field a distance r away from a long wire with charge density π. π π¬=± π₯ 2π • This one is especially important. • This is the electric field anywhere away from a large sheet of charge. • Notice that the electric field doesn’t depend on distance, and always points perpendicular to the surface. • Classic problem: What is the electric field between two parallel plates of charge ±Q and area A? • Answer: Since you have two plates of opposite charge, the E fields add, and thus… • π= π π΄ πΈ=2 π 2π0 = π π0 = π π΄π π΄ π → π: π → π¬: ππ − ππ = − π¬ ⋅ ππ π΅ π¬ = −πV • These relations let you go from either E to V or vice-versa. If you know one, you can calculate the other. π = −Δπ = −π0 Δπ This equation can be used to: • If you’re given Δπ, you can find the work done. • If a point charge of charge q0 goes through a potential difference of Δπ, it tells you the work done on the charge. π = π0 π • This is the defining relation between potential energy (U) and electric potential (V). • Note: since q0 can be positive or negative, U and V do not necessarily have the same sign. • One more time: Electric potential (V) is not the same thing as electric potential energy (U) • But let’s rewrite it. π = π/π0 Similar to: π¬= π . π0 ππ π π= π • Electric potential of a point charge q, a distance r away, assuming V=0 at infinity. • Potential goes up as you get closer to the point charge. Δπ = ±πΈπ • Rather important: This is the potential difference between two parallel conducting plates, otherwise known as a capacitor. • What is E for a capacitor again? πΈ = π/π0 π πΆ= π • Definition of capacitance • C=Charge Q/ Voltage drop π΄ πΆ = π0 π • Special case of capacitance when you’re looking at a parallel plate capacitor. • Notice that the capacitance doesn’t depend on the charge on the plates. πΆππ = πΆ1 + πΆ2 + β― • Adding capacitors in PARALLEL. 1 1 1 = + +β― πΆππ πΆ1 πΆ2 • Adding capacitors in SERIES. ππ πΆ = 4ππ0 π−π • Special case of capacitance when you’re looking at a two concentric spherical conducting shells. • The radius of the smaller shell is a, the radius of the larger shell is b. πΆ = 4ππ0 π • The capacitance of a single spherical shell of radius R 2 1 2 π 1 π = πΆπ = = ππ 2 2πΆ 2 • You use these equations to calculate the stored energy in a capacitor • Okay, but there’s 3 different equations, so which one is appropriate for my problem? – If they just ask you to calculate U, use the one that has variables you know. – If they ask you what happens to U if you double the charge, halve d, etc… • If the capacitors are stand-alone, use π2 . 2πΆ 1 2 • If the capacitors are connected to a voltage source, use πΆπ 2 π = π π0 • This relates the permittivity of free space to the permittivity in a medium of dielectric constant π . πΆ = π C0 • Relates capacitance without a dielectric (C0) to capacitance with a dielectric. 1 2 π’ = ππΈ 2 • The energy density when you have an electric field E in a medium of permittivity π. • For example, let’s say you have a cube (L=3) filled with water (π = 78). The cube has the same E field everywhere (E=5). – Then 1 u= 2 ∗ 78 ∗ π0 ∗ 52 3 1 2 – Total stored energy=π = ∫ π’ ππ = 3 ⋅ ⋅ 78 ⋅ π0 ⋅ 52 ππ πΌ= ππ‘ • Definition of current. π = πΌπ • Ohm’s Law: A relationship between voltage, current, and resistance. • Pretty fundamental. πΌ π½ = = πππ£π π΄ • π½=Current density • n = density of charge carriers • Vd=drift velocity (average velocity of charge carriers) • q=charge on a charge carrier (usually e=1.6 ∗ 10−19 πΆ) πΈ = π½π • Microscopic Ohm’s Law: • E-field (E) = Current density (J) x resistivity (π) πΏ π =π π΄ • Resistance of a conductor of resistivity π, length L, and cross-sectional area A π = π0 1 + πΌ π − π0 • Resistivity changes as a function of temperature. π = π 0 1 + πΌ π − π0 • Resistance changes as a function of temperature π ππ = π 1 + π 2 + β― • Adding resistances in series. 1 1 1 = + +β― π ππ π 1 π 2 • Adding resistances in parallel. Practice Problems In the figure to the right, there are two charges connected by a massless insulating rod…and remember to use VECTORS when appropriate… Draw the electric dipole. Torque caused by the electric field= Dipole moment= Potential energy as it is right now= Which way will the dipole begin to rotate? (Clockwise/Counter-clockwise) How much work is done in rotating the dipole from its current position to the stable equilibrium position? What does the work in question f? A block of MagicFoam (length 10 cm, width 10 cm, height 3 cm) sits on top of a calm body of water. MagiFoam density=0.5 g/cm3. How much of the block is submerged? A 10 kg block floats in the water. What is the buoyant force on it? A house with a roof of area 5 m2 has winds of 50 m/s above it. What is the force on the roof caused by the pressure difference? How much energy does it take to bring two electrons within .1 nπ of each other? Some water is flowing at a rate of 20 mph in a pipe. Further on in the pipe, the pipe halves its diameter. What is the new speed of the water? If current is going from your hand to your foot, which direction are the electrons going? Final Questions? Thank you, and good luck!