Chapter 16 Capacitors Batteries Parallel Circuits Series Circuits

Hint: Be able to do the homework (both the problems to turn in AND the recommended ones) you’ll do fine on the exam! Friday, February 26, 1999 in class Ch You may bring one 3”X5” index card (hand-written on both sides), a pencil or pen, and a scientific calculator with you.

-Q A common type of capacitor consists of a pair of parallel conducting plates, one charged positively, one charged negatively. d +Q

-Q Therefore…. An electric field exists between the plates of a capacitor. d E +Q

Capacitance is the ability to store charge. On what could capacitance possibly depend? d Think geometrically…. C =  o A/d A

 o = permittivity of free space = 8.85 X C 2 /Nm 2 This quantity is related to another constant with which we are already familiar in electrostatics: k = 1 / 4  o

C =  o A/d [ C ] = [  o ] [A] / [d] C 2 m 2 [ C ] = N m 2 m C 2 = J C = V 1 Farad = C / V

What happens when you connect a capacitor to a battery? + _ V C + _ Circuit Diagram

The battery converts internal chemical energy to electrical energy, pulling electrons off the top plate of the capacitor and pushing them onto the lower plate of the capacitor until the capacity of the capacitor is reached. At that point, each plate of the capacitor holds a charge Q. The battery maintains a potential difference across its terminals (and hence, the capacitor) of V. V C + _ +Q - Q

C = Q / V Notice this has the same units as the quantity we derived earlier! 1 Farad = C / V

C = Q / V This definition of capacitance is particularly useful since it does not require us to have any knowledge about the geometry of the capacitor. Using this definition, capacitance can be determined solely from the behavior of the electrical circuit.

V C2C2 + _ C1C1 So, what happens after the battery is connected to this circuit? Charge of Q 1 = C 1 V accumulates on capacitor C 1. Charge of Q 2 = C 2 V accumulates on capacitor C 2.

V C2C2 + _ C1C1 We can construct an equivalent circuit with a single capacitor... V C eq + _ +(Q 1 +Q 2 ) - (Q 1 +Q 2 ) C eq = (Q 1 + Q 2 ) / V = C 1 + C 2 Capacitors in parallel ADD.

So, what happens after the battery is connected to this circuit? V 1 + V 2 = V V C2C2 + _ C1C1 +Q-Q +Q -Q Q 1 = Q 2 due to conservation of charge!

V C 2, V 2 + _ C 1, V 1 +Q-Q +Q -Q We can construct an equivalent circuit with a single capacitor... V C eq + _ +Q - Q C eq = Q / V V 1 = Q / C 1 V 2 = Q / C 2 V = V 1 + V 2

Capacitors in series ADD INVERSELY. V C eq + _ +Q - Q C eq = Q / V Q C eq = Q / C 1 + Q / C = + C eq C 1 C 2

_ +  V = 1 Volt e-e- An electron is accelerated across a potential difference of 1 Volt. It gains kinetic energy as it moves across the potential, losing potential energy. This particular amount of energy is known as...

The amount of energy 1 electron gains when accelerated through a potential difference of 1 Volt. 1 e V = (1.6 X C) (1 V ) = 1.6 X J

Capacitors store charge. It takes work to put charges on capacitors. That work becomes the potential energy of the capacitor. So capacitors store energy. We take advantage of this all the time! Can you name some examples of devices that use the energy stored in capacitors?

How much work does it take to charge a capacitor? W = q V Start with uncharged plates. V =0 So it requires almost no work to bring up the first bits of charge,  q. +q+q -q-q

+q+q -q-q Now that our capacitor has a charge  q, what is the potential difference between the plates? V = Q / C As we bring up more and more charge, V increases with Q at the rate 1/ C, so we can plot V as a function of Q: V Q Slope = 1 / C qq

The total amount of work to bring charge Q onto an initially uncharged capacitor, therefore, is simply the area under the curve: V Q V = Q / C Q = C V