EXTRA SI EXAM REVIEW Tuesday 12-2 pm Regener 111 Along with regular SI at 3-4. HW due Wednesday 24-46, (also MP online) It is proposed to store solar energy by filling a water tank during the day, raising the water from ground level. (At night, the water drains turning a generator.) A tank is 20 m high and has vertical sides. When half full, it stores 1 million Joules. How much energy will it store when full? A] 1 million Joules B] 2 million Joules C] 4 million Joules D] 20 million Joules

“Dielectric” = a fancy word for a real insulator. In real insulators, charges can move, but only very small distances… the electrons can shift a bit with respect to the nucleii, but they can’t hop from one nucleus to the next. What happens to the electric field in a capacitor if we add a dielectric material? Let’s sketch on the board, and then I’ll quiz you.

By superposition, the field in the dielectric is the sum of the original capacitor field, plus the field from the polarization surface charge. The total field is therefore A] bigger than the original capacitor field B] smaller than the original capacitor field C] the same as the original capacitor field

Answer: smaller than the original field. We can write that the new field is the original field, divided by K, the “dielectric constant” What is the new potential across the capacitor? A] same as the old potential, without the dielectric B] bigger than the old potential, by a factor of K C] smaller than the old potential, by a factor of K

If I put a charge ±Q on a parallel plate capacitor and insert a dielectric, what happens to V? A] it increases B] it decreases C] it stays the same

If I put a charge ±Q on a parallel plate capacitor and insert a dielectric, what happens to V? The field in the dielectric is reduced by a factor of K. So the potential goes down. What happens to the potential energy? A] it goes up by a factor of K B] it goes up by a factor of K 2 C] it goes down by a factor of K D] it goes down by a factor of K 2 E] it stays the same.

If I put a charge ±Q on a parallel plate capacitor and insert a dielectric, what happens to V? What happens to the potential energy? It may be counterintuitive, but the potential energy goes down by a factor of K (not K 2 ). U=Q 2 /(2C); the capacitance goes up by a factor of K. (or U = CV 2 /2 … the voltage drops by K, the capacitance increases by K) Note: the energy density for a given E field in a dielectric is The field goes down by a factor of K, but epsilon adds a factor of K.

If the potential energy goes down, where does the energy go? The field pulls the dielectric into the capacitor, giving it kinetic energy. (Very small, here.) Application: Optical tweezers

With a fixed charge on the plates, the dielectric is pulled into the capacitor. The field and field energy go down. What happens if, instead, the capacitor is held at a fixed potential? The energy goes up! U = CV 2 /2, and the capacitance goes up. But we cannot conclude that you must push the dielectric in (in fact, you don’t have to) … the extra energy comes from the battery.

Batteries don’t store charge. They store energy. A chemical battery works because electrons like to leave some materials to go to others. The change in energy when the electron goes “downhill” becomes available as electric POTENTIAL between the battery terminals

Lead Acid (Car) Battery The rxn on the right will not go in the direction indicated unless the electrolyte sol’n potential is closer than V to the + electrode potential. i.e. the + electrode can be no more than V higher in potential than the electrolyte sol’n. Pb +2 is in the form of solid lead sulfate on the electrodes. When “discharged”, both electrodes turn into lead sulfate.

Lead Acid (Car) Battery The rxn on the left will not go in the direction indicated unless the electrolyte potential is closer than V to the - electrode potential. i.e. the - electrode can be no more than V lower in potential than the electrolyte soln.

Lead Acid (Car) Battery Overall, then, the reactions will STOP when the potential between the +/- electrodes is 2.041V. Only a tiny tiny tiny amount of charge needs to build up on the electrodes for the reaction to stop. How much? But if you connect the electrodes with a resistive wire, the reaction will start to go as the potential drops a hair below 2.041V.

Lead Acid (Car) Battery What happens if you force the potential difference to be higher than V? The reactions run backwards! Lead sulfate turns into lead oxide and lead (metallic). This is charging the battery.

Lead Acid (Car) Battery Note that the rxn doesn’t make a big “reservoir” of electrons. The battery doesn’t die because it runs out of stored electrons. It doesn’t store electrons. It “pumps” electrons “on demand”, i.e. when the potential falls below V.

Exam scoring Everyone gets an extra point because Dr. Thomas thinks 4 2 =4. With your extra point, 13 is the lowest C 15 is the lowest B- 17 is the lowest A- Note: Just because you can find the flux through the side of a cube doesn’t mean that the field there is flux  area! If you do better on this material on the final exam, I will substitute that part of your final grade for your grade on this exam.

Batteries: A pump for charges, creating a tiny tiny tiny reservoir of charge at high potential. When connected to a circuit, the charge flows (we call this a current). It would be all gone in a nanosecond, but as soon as the potential drops a hair, the chemical reaction pumps MORE CHARGE. A typical car battery has a few nanoCoulombs of charge built up. When running, it sends 20 COULOMBS of charge PER SECOND (20 amperes) through the circuit. Batteries store energy, not charge.

Electric fields in Conductors. In electrostatics, there can be no field in a conductor. The charges move until there is no field left. But if we have a battery, we can keep the charges moving indefinitely (until the battery runs out of ENERGY.) So in circuits, we can (and do) have fields in conductors.

If we have a charge in a electric field, we expect it to accelerate. Since current is the charge per unit time (passing a point), we might expect the current to increase with time. It does not. If we apply a steady field across a conductor, (by pumping charges with a battery), the current is steady. This happens because the mobile charges (electrons, in metal) collide with defects and “misplaced” atoms in the metal. (Remarkably, mobile charges do NOT collide with “perfectly placed” atoms. This is a quantum mechanical effect!)

If there is an average time  between collisions, it is easy to show that the electrons have a net “drift” velocity in the field given by So how much current is there? And what is the “current density” (current per unit area)?

We see that current density is proportional to E. This is called Ohm’s Law (microscopic version) We define resistivity as the reciprocal of the proportionality constant. (Conductivity = 1/resistivity) These are material properties. Different materials have different resistivity.

Since current = current density x area, we can find a relation between current and voltage drop for a resistor (conductor) We combine the geometric terms with the resistivity to get a proportionality constant we call resistance, R. V = IR And R =  L/A

The two electrodes of an ideal V volt battery are shown. The field lines for the electric field between the electrodes are shown when nothing is attached to the electrodes. The electrodes have a separation = d. A resistive wire of length L and uniform diameter is then attached to the battery. What is the electric field in the wire, when steady state is reached? (It takes about a nanosecond or so for steady state to be reached.) A] 0 B] it varies, but averages to V/d C] it varies, but averages to V/L D] it is V/L everywhere in the wire E] no way to determine

Where on this wire will negative charges “build up” (a tiny amount) A B Somehow, the wire has changed the electric field so that it is uniform in magnitude and directed along the wire. It can only do this because there are tiny tiny tiny tiny amounts of charge that build up on the wire.