Shielding: The electric field inside conductors

Shielding: The electric field inside conductors

This section examines how, in any electric field at all, people can make a little safe haven a region of zero electric field with only the aid of a hollow conductor! Figure 3-8 shows an internal cross-section of a spherical conducting ball. Say that some charges were planted inside the solid metal ball. Now, electric charges always generate an electric field, and conducting materials let charges flow freely in response to electric field, so what happens? Figure 3-8: Charges inside a conductor. The electric field generated by the charges implanted in the conducting material push other, similar charges away.

Getting the lowdown on electric potential

If you have a mass in a gravitational field, it has potential energy. As you throw a ball upward, for example, the kinetic energy of its motion is converted to gravitational potential energy as it reaches its peak, and then the potential energy changes back to kinetic energy as the ball falls back to you. The gravitational forces on the ball do work and exchange potential and kinetic energy. Because a force likewise acts on charges in an electric field, you can speak about potential energy here, too. That potential energy is electric potential energy. What makes all forms of energy essentially the same is that they can all be converted to mechanical work. As you may remember from Physics I, work done (W) is the result of a force (F) moving a body through a distance (s), and they’re related thus: W = Fs. The energy in the interaction of a charge with an electric field is converted to work when the electrical forces move the charge. Moving twice as much charge takes twice as much work for the electrical forces. The work that’s done for every unit of charge is the voltage. In physics, voltage is called electric potential (not electric potential energy, which isn’t per unit of charge); sometimes, it’s just shortened to potential. Instead of using the term voltage, it’s more correct to say that electric potential is measured in volts, whose symbol is V

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Finding the electric potential from charges

Say you have a point charge Q. What’s the electric potential due to Q at some distance r from the charge? You know that the size of the force on a test charge q due to the point charge is equal to the following (see the earlier section “Looking at electric fields from charged objects” for details): where k is a constant equal to 8.99 × 109 N-m2/C2, Q is the point charge measured in coulombs, q is the charge of the test charge, and r is the distance between the point charge and the test charge. You also know that the electric field at any point around a point charge Q is equal to the following: Thus, close to the point (when r is small), E is large and the field is therefore strong. As you move away from the point charge, r increases and the electric field quickly becomes weak. Suppose you place a small test charge, q, in this field and try moving it around. The test charge feels a strong force close to the point charge, which quickly falls away as you move it to a greater distance. If the test charge is of the opposite sign to the point charge, you have to do work to pull it away from the point charge. This means that the test charge has a lower potential energy closer to the point charge (this is reversed if the charges are the same sign). So what’s the electric potential from the point charge? At an infinite distance from the point charge, you can’t see it or be affected by it, so set the potential from the point charge to be zero there. As you bring a test charge closer, to a point r away from the point charge, you have to add up all the work you do and then divide by the size of the test charge. And the result after you do turns out to be gratifyingly simple. Here’s what that it looks like: So the electrical potential is large close to the point charge (when r is small) and falls away at greater distances. This idea applies to all point charges, so what does it mean for the electrons orbiting the protons in an atom? How hard do you have to work to pull an electron away from an atom? First find the electrical potential. The size of the charge of the electron and the proton is 1.6 × 10–19 coulombs. The electron and proton are typically 5.29 × 10–11 meters apart, so the electric potential is That close to the proton, the electric potential is a full 27.2 volts. That’s quite something for such a tiny charge! The amount of energy needed to move an electron through 1 volt is called an electron-volt (eV). So you may expect that you’d need 27.2 eV of energy to pull an electron out of the atom. But the electron is moving quickly, so it already has some energy to contribute. In fact, the electron has so much kinetic energy that you need only half that, 13.6 eV of energy, to win the electron from the atom!

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Illustrating equipotential surfaces for point charges and plates

To illustrate electric potential, you can draw equipotential surfaces; that is, surfaces that have the same potential at every point. Drawing an equipotential surface gives you an idea of what the electric potential from a charge or charge distribution looks like. For example, on one equipotential surface, the potential could always be 5.0 volts or 10.0 volts. Because the potential from a point charge depends on the distance you are from the point charge, the equipotential surfaces for a point charge are a set of concentric spheres. You can see what they’d look like in Figure 3-11.

In other words, the equipotential surfaces here depend only on how far you are between the two plates. You can see this in Figure 3-12, which shows two equipotential surfaces between the plates of the parallel plate capacitor. Tip: This is analogous to the gravitational potential close to the ground, which increases in proportion to height. They’re both cases of uniform fields in which the field is the same at every point.

Hold on tight: Finding the binding energy of the nucleus

The strong force is what holds nuclei together — which means that separating the nucleons in a nucleus takes work. That work is called the binding energy of the nucleus. How do you find the binding energy of a nucleus without taking it all apart? You can be clever about this and measure a nucleus’s mass compared to the masses of its constituent nucleons. That is, when you find the mass of the nucleus, the nucleus is less massive than the sum of the nucleons that go into it (so you could say that a nucleus is less than the sum of its parts). Why? Because some mass went into the binding energy of the nucleus. The difference between the masses of all the nucleons separately and the final nucleus is called the mass defect of the nucleus, which has the symbol Δm. So the mass defect of a nucleus is where Σmnucleons is the sum of the masses of the nucleons and mnucleus is the mass of the nucleus after all the nucleons are put together. So how can you find the binding energy from the mass defect? You may recall that Einstein said that E0 = mc2, so the binding energy of a nucleus is equal to Ebinding = Δmc2 where Δm is the mass defect of the nucleus. Finding the mass defect Check out some numbers. For example, grab a standard helium atom: . The nucleus of that atom has two protons and two neutrons, and experiments show that it has a mass of 6.6447 × 10–27 kilograms. What is its mass defect? You can find the mass defect of a nucleus with . Here, that becomes ΔmHe-4 = 2mproton + 2mneutron – mnucleus Like any good physicist, you know that ✓ The mass of a proton is 1.6726 × 10–27 kilograms ✓ The mass of a neutron is 1.6749 × 10–27 kilograms 22_538067-ch15.indd 327 6/1/10 10:21 PM 328 Part IV: Modern Physics So you have the following: ΔmHe-4 = 2(1.6726 × 10–27 kg) + 2(1.6749 × 10–27 kg) – (6.6447 × 10–27 kg) ≈ 0.0503 × 10–27 kg = 5.03 × 10–29 kg That’s very small indeed. Calculating the binding energy What’s the binding energy of the standard helium atom? That equals Ebinding = Δmc2. The mass defect, m, is 5.03 × 10–29 kilograms, and the speed of light is approximately 3.00 × 108 meters per second, so you have the following: Ebinding = (5.03 × 10–29 kg)(3.00 × 108 m/s)2 ≈ 4.53 × 10–12 J What’s that in electron-volts, eV? An electron-volt is the energy needed to push one electron through 1 volt of electric potential, and 1 electron-volt is 1.60 × 10–19 joules, so the binding energy of He-4 is So the binding energy is 28.3 million electron-volts. And because the proton has the same charge as the electron, 28.3 million electron-volts is the energy you get by letting a proton drop through 28.3 million volts. Put another way, it takes 24.6 eV to pull an electron away from an atom of He-4. It takes more than 1 million times that amount to pull a proton out of the He-4 nucleus. That’s the strong force at work — and it has to overcome the repulsive force of two protons at extremely close distance as well.

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