---

	A capacitor can be "charged" and can store charge. When a capacitor is being charged, negative charge is removed from one side of the capacitor and placed onto the other, leaving one side with a negative charge (-q) and the other side with a positive charge (+q). The net charge of the capacitor as a whole remains equal to zero.
Fig. 24.1 Any two conductors insulated from one another form a CAPACITOR.

CAPACITOR     Fig. 24.2 (a) A charged parallel plate capacitor. (b) When the plate separa- tion is small compared to the plate area, the fringing of the electric field at the edges is negligible.	The amount of charge that can be placed on a capacitor is proportional to the voltage pushing the charge onto the positive plate. The larger the potential difference (voltage) between the plates, the larger the charge on the plates:                   Q = C V The constant of proportionality is called the "capacitance" and is proportional to the area (A) of one of the plates and inversely proportional to the separation between the plates (d):                 C = e A / d for a parallel plate capacitor, where e is the permittivity of the insulating material (or DIELECTRIC) between the plates. Recall that we used Gauss's Law to calculate the magnitude of the electric field (E) between the plates of a charged capacitor: E = s / eo where the space between the plates is a vacuum. Vab = E d, so E = Vab / d

Fig. 24.6 (a) Two capacitors in series and (b) The equivalent capacitor.

Fig. 24.7 (a) Two capacitors in parallel and (b) the equivalent circuit.

24.3 ENERGY STORAGE in CAPACITORS and ELECTRIC FIELD ENERGY Capacitors can store charge and ENERGY. We can calculate the potential energy (U) stored in a charged capacitor from the ideas we learned in the last chapter (Electric Potential). Since the charging of a capacitor can be thought of as moving charge from one plate DIRECTLY onto the other plate through a potential difference of V, the voltage between the capacitor plates. Recall: DU = q DV, and the potential V increases as the charge is placed on the plates (V = Q / C). Since the V changes as the Q is increased, we have to integrate over all the little dq being added to a plate: DU = q DV leads to U = ò V dq = ò q/c dq = 1/C ò q dq = Q^2 / 2C. And using Q = C V, we get U = Q^2 / 2C = 0.5 C V^2 = 0.5 Q V So the energy stored in a capacitor can be thought of as the potential energy stored in the system of positive charges that are separated from the negative charges, much like a stretched spring has potential energy associated with it.     ELECTRIC FIELD ENERGY Here's another way to think of the energy stored in a charged capacitor. If we consider the space between the plates to contain the energy (equal to 1/2 C V^2) we can calculate an energy DENSITY (Joules per volume). The volume between the plates is Area x distance between plates, or A d. Then the energy density (u) is u = 1/2 C V^2 / A d = 1/2 eo E^2 where we have used C = Eo A / d and V =E d. This is an important result because it tells us that empty space can contain energy, i.e., if there is an electric field in the "empty" space. If we can get an electric field to travel (or propagate) through empty space we can send or transmit energy!!!

Fig. 24.12 Effect of a dielectric between the plates of a parallel plate capacitor. The electrometer measures the potential difference. (a) With a given charge, the potential difference is Vo. (b) With the same charge but with a dielectric between the plates, the potential difference V is smaller than Vo.

Most capacitors have an insulating (or DIELECTRIC) material between the plates. The presence of this dielectric INCREASES the capacitance of the capacitor compared to when the space between the plates was empty (a vacuum). Here's why: From measurements like the one above we know that insertion of a dielectric between the plates of a capacitor causes the potential difference Vo between the plates to decrease to V. The original capacitance, from Q = C V, is given by Co = Q / Vo and since Q must stay constant (it has nowhere to go!) we can write Q = Q, or Co Vo = C V, and if Vo decreases to V, Co must increase to C to keep the equation balanced. We can now define K = C / CO = "dielectric constant" = the ratio of the capacitances. And from Co Vo = C V (above) we can write V = Vo Co / C = Vo / K. The dielectric constant (K) is a positive number equal to 1 for a vacuum and greater than 1 for other dielectric materials (Teflon - 2.1, glass = 7, water = 80, etc.), so we can say that the potential of a capacitor decreases by a factor of K when a dielectric material is added and the charge on the capacitor stays constant (i.e., the capacitor is not connected to a battery while the dielectric is inserted).

And since V = E d, we can conclude that the electric field between the plates must be reduced as a dielectric is inserted Q = (constant). See below for a physical picture of what is happening.

Fig. 24.13 (a) Electric field lines with vacuum between the plates. (b) The induced charges on the faces of the dielectric decrease the electric field. (See figure below for a physical picture of how the (blue) induced charges on the faces of the dielectric are established.)

The electric field between the capacitor plates is reduced by the presence of the dielectric because the induced surface charges on the dielectric (see figure below) cause an electric field in the opposite direction of the original field in the charged capacitor. These fields tend to cancel each other resulting in a reduction of the original field.

	Molecules in the dielectric material have their positive and negative charges separated slightly, causing the molecules to be oriented slightly in the electric field of the charged capacitor.
Fig. 24.18 Polarization of a dielectric in an electric field gives rise to thin layers of bound charges on the surfaces, creating positive and negative surface charge densities. The sizes of the molecules are greatly exaggerated for clarity.

Fig. 24.19 (a) Electric filed of magnitude Eo between two charged plates. (b) Introduction of a dielectric of dielectric constant K. (c) the induced surface charges and their field (thinner lines). (d) Resultant filed of magnitude Eo/K when a dielectric is between charged plates.

---

It takes time to charge a capacitor and it takes time to discharge one. This time is dependent on the sizes of the capacitor and the resistor in the circuit. The product of RC has units of seconds. The case for charging a capacitor is described first, then discharging a capacitor.

Fig. 26.20 CHARGING a capacitor. Just before the switch is closed the charge on the capacitor is zero. When the switch is closed (at time t= 0) the current jumps from zero to:   i = e / R.

	For a CHARGING capacitor, I(t) = dq/ dt = e / R exp (- t/RC) = Io exp (- t/RC)         q(t) = Ce [1 - exp (- t/RC)] = Qf [1 - exp ( - t/RC)]       Both these functions are plotted to the left.
Fig. 26.21 CHARGING capacitor

Fig. 26.22 DISCHARGING a capacitor. Just before the switch is closed (at t= 0) the charge on the capacitor is Qo, and the current is zero. At the time t after the switch is closed, the charge on the capacitor is q, the current is i.

	For a DISCHARGING capacitor, I(t) = dq/ DT = - Qo / RC exp (- t/RC)       q(t) = Qo exp (- t/RC)         Both these functions are plotted to the left.
Fig. 26.23 DISCHARGING capacitor