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A DC cell creates a potential difference between its terminals by transferring electrons from the positive to the negative terminal. As it transfers more and more electrons, the negative terminal gains a greater charge, and the terminal exerts a greater force on every additional electron the cell transfers. Thus it takes more energy to transfer each additional electron onto it. Finally, the energy cost of adding an electron is too much and the cell stops, but by that time the terminal potential has been reached (Figure 15-23).

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Figure 15-23


If, however, we connect each terminal to a plate of metal (Figure 15-24), then the cell is able to transport more electrons at a low energy cost, because the additional electrons are able to spread out across the plate. Eventually the cell tries to push one more electron onto the plate but cannot, because the energy cost is too great. In this case the potential between the plates is same as before, but the difference is that more charge is able to be contained on the plates than on the terminals of the cell.

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Figure 15-24

If we bring the plates near each other but not touching (Figure 15-25), even more charge is able to be transferred through the cell. An electron arriving at the negative plate feels the opposing force from the electrons, but it also feels the attractive force of the nearby positive plate, so it does not mind so much getting onto the negative plate. (That is, it can be energetically favorable.)
Such a device, which holds a charge when a potential difference is applied to it, is called a capacitor. In Figure 15-25, if a larger potential is applied to the plates, a proportionally larger charge will sit on them. The capacitance, the capacity to hold charge, is defined by


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Figure 15-25

where Q is the charge on one plate, and AV is the potential difference across the plates. The units of capacitance are [Coulombs/volt = Farads = F).


To reiterate, the capacitance of a device depends on how it is built. The charge on it is proportional to the voltage applied to it: Q = CV.

We place two metal plates parallel to each other and 0.01 meters apart. One sheet we connect by a wire to the positive end of a 6-V DC cell; and the other, to the negative end. The two plates now have a 6-V potential difference between them. The electric field between the plates is uniform (though we will not prove it), that is, the same direction and magnitude everywhere. A dust mite has a 2 X 10–14 C charge tied around his ankle.

  1. How much work is required for him to cross from the negative plate to the positive plate?
  2. What force does he experience as he crosses?
  3. What is the magnitude of the electric field between the plates?
  1. First let’s DRAW A DIAGRAM showing the electric field (Figure 15-26); second, a diagram showing the forces on the mite (Figure 15-27).
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Figure 15-26 Figure 15-27

Moving from a negative to a positive plate is an uphill battle for the mite, so we expect the work to be positive. We write
W = qΔV
= (2 x 10–14 C)(6 J/C)
= 1.2 x –13 J
  1. If the electric field is uniform, then the force (Felec = qE) that the mite experiences must be constant. If he goes straight across, then we have the old work equation
W = FmiteΔxcosφ
Now the mite moves at a constant velocity, so the forces are balanced and we can replace Fmite with Felec. Also, we know cosφ is 1. Thus we write
W = FelecΔx
Felec = Wx
= (1.2 x 10–13J)/0.01m = 1.2 x 10–11 N
  1. We obtain the electric field from Felec = qE, so that E = 600 N/C.
It is more important to understand the pictures and the ideas than to apply the formulas in this example.

The equation we derived is worth remembering in its own right:


where ΔV is the potential across a capacitor. E is the magnitude of the electric field inside of the capacitor, and Δx is the separation of the plates.


Remember what this equation is about. This is just our old W = F∆x equation in new clothing, that is, work (per charge) is force (per charge) times displacement. In fact, the assumption made in the problem is fairly accurate: The electric field between two charged parallel plates is uniform.
Now let’s complicate the situation even more. A dielectric is a nonconducting substance, such as plastic. Its electrons are not free to move from one atom to the next, but the electrons can slosh a little onto one side of the molecules (they are slightly polarizable). If a dielectric is placed between the plates, then the molecules in the dielectric become polarized, with electrons being pulled toward the positive plate (Figure 15-28). The result is that the DC cell can transfer still more charge from one plate to the other, since an electron arriving at the negative plate feels also the slight positive charge of the one side of the dielectric. Thus, placing the dielectric between the plates increases the capacitance.

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Figure 15-28

Each material which is a nonconductor has a dielectric constant κ. The capacitance of the capacitor with the dielectric in the middle is given by


Cdielectric = κCvacuum…. (10)


The constant κ is always greater than 1.

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