How much stored energy is there between the spheres

The space between its plates has a volume Ad , and it is filled with a uniform electrostatic field E. The total energy of the capacitor is contained within this space.

The energy density in this space is simply divided by the volume Ad. If we know the energy density, the energy can be found as. If we multiply the energy density by the volume between the plates, we obtain the amount of energy stored between the plates of a parallel-plate capacitor:.

In this derivation, we used the fact that the electrical field between the plates is uniform so that and Because , we can express this result in other equivalent forms:.

The expression in Figure for the energy stored in a parallel-plate capacitor is generally valid for all types of capacitors. To see this, consider any uncharged capacitor not necessarily a parallel-plate type. At some instant, we connect it across a battery, giving it a potential difference between its plates. Initially, the charge on the plates is As the capacitor is being charged, the charge gradually builds up on its plates, and after some time, it reaches the value Q.

To move an infinitesimal charge dq from the negative plate to the positive plate from a lower to a higher potential , the amount of work dW that must be done on dq is. This work becomes the energy stored in the electrical field of the capacitor.

In order to charge the capacitor to a charge Q , the total work required is. Since the geometry of the capacitor has not been specified, this equation holds for any type of capacitor. The total work W needed to charge a capacitor is the electrical potential energy stored in it, or.

When the charge is expressed in coulombs, potential is expressed in volts, and the capacitance is expressed in farads, this relation gives the energy in joules. Knowing that the energy stored in a capacitor is , we can now find the energy density stored in a vacuum between the plates of a charged parallel-plate capacitor. We just have to divide by the volume Ad of space between its plates and take into account that for a parallel-plate capacitor, we have and. Therefore, we obtain.

We see that this expression for the density of energy stored in a parallel-plate capacitor is in accordance with the general relation expressed in Figure. We could repeat this calculation for either a spherical capacitor or a cylindrical capacitor—or other capacitors—and in all cases, we would end up with the general relation given by Figure. Energy Stored in a Capacitor Calculate the energy stored in the capacitor network in Figure a when the capacitors are fully charged and when the capacitances are and respectively.

Strategy We use Figure to find the energy , , and stored in capacitors 1, 2, and 3, respectively. The total energy is the sum of all these energies. Solution We identify and , and , and The energies stored in these capacitors are. Significance We can verify this result by calculating the energy stored in the single capacitor, which is found to be equivalent to the entire network.

The voltage across the network is The total energy obtained in this way agrees with our previously obtained result,. Check Your Understanding The potential difference across a 5. By what factor is the stored energy increased? In a cardiac emergency, a portable electronic device known as an automated external defibrillator AED can be a lifesaver. A heart attack can arise from the onset of fast, irregular beating of the heart—called cardiac or ventricular fibrillation.

Today, it is common for ambulances to carry AEDs. AEDs are also found in many public places. These are designed to be used by lay persons.

CPR cardiopulmonary resuscitation is recommended in many cases before using a defibrillator. Capacitance of a Heart Defibrillator A heart defibrillator delivers of energy by discharging a capacitor initially at What is its capacitance?

Strategy We are given and V , and we are asked to find the capacitance C. We solve Figure for C and substitute. Solution Solving this expression for C and entering the given values yields. If you wish to store a large amount of energy in a capacitor bank, would you connect capacitors in series or parallel?

How much energy is stored in an capacitor whose plates are at a potential difference of 6. Best regards, Mike Gottlieb mg feynmanlectures. In the study of mechanics, one of the most interesting and useful discoveries was the law of the conservation of energy. The expressions for the kinetic and potential energies of a mechanical system helped us to discover connections between the states of a system at two different times without having to look into the details of what was occurring in between.

We wish now to consider the energy of electrostatic systems. In electricity also the principle of the conservation of energy will be useful for discovering a number of interesting things.

The law of the energy of interaction in electrostatics is very simple; we have, in fact, already discussed it. There is some energy in the system, because a certain amount of work was required to bring the charges together. We have already calculated the work done in bringing two charges together from a large distance.

It follows, therefore, that the total energy of a system of a number of charges is the sum of terms due to the mutual interaction of each pair of charges. We shall concern ourselves with two aspects of this energy. One is the application of the concept of energy to electrostatic problems; the other is the evaluation of the energy in different ways. Sometimes it is easier to compute the work done for some special case than to evaluate the sum in Eq.

As an example, let us calculate the energy required to assemble a sphere of charge with a uniform charge density. The energy is just the work done in gathering the charges together from infinity. Imagine that we assemble the sphere by building up a succession of thin spherical layers of infinitesimal thickness. We can also interpret Eq. We consider now the energy required to charge a condenser. How much work is done in charging the condenser?

We now consider applications of the idea of electrostatic energy. Consider the following questions: What is the force between the plates of a condenser? Or what is the torque about some axis of a charged conductor in the presence of another with opposite charge? Such questions are easily answered by using our result Eq.

This work must be equal to the change in the electrostatic energy of the condenser. By Eq. It is easy to see how the idea is extended to conductors of any shape, and for other components of the force. In Eq. We could, in this way, find the torque on the movable plates in a variable condenser of the type shown in Fig.

But we have a surprising factor of one-half. If we imagine that the charge at the surface of the plate occupies a thin layer, as indicated in Fig. That is why the factor one-half is in Eq. You should notice that in computing the virtual work we have assumed that the charge on the condenser was constant—that it was not electrically connected to other objects, and so the total charge could not change.

Suppose we had imagined that the condenser was held at a constant potential difference as we made the virtual displacement. Also, we know that two plates with opposite electrical charges must attract. The principle of virtual work has been incorrectly applied in the second case—we have not taken into account the virtual work done on the charging source.

We now consider an application of the concept of electrostatic energy in atomic physics. We cannot easily measure the forces between atoms, but we are often interested in the energy differences between one atomic arrangement and another, as, for example, the energy of a chemical change. Since atomic forces are basically electrical, chemical energies are in large part just electrostatic energies. An ionic crystal like NaCl consists of positive and negative ions which can be thought of as rigid spheres.

They attract electrically until they begin to touch; then there is a repulsive force which goes up very rapidly if we try to push them closer together. For our first approximation, therefore, we imagine a set of rigid spheres that represent the atoms in a salt crystal.

The structure of the lattice has been determined by x-ray diffraction. It is a cubic lattice—like a three-dimensional checkerboard. Figure 8—5 shows a cross-sectional view. If our picture of this system is correct, we should be able to check it by asking the following question: How much energy will it take to pull all these ions apart—that is, to separate the crystal completely into ions? This energy should be equal to the heat of vaporization of NaCl plus the energy required to dissociate the molecules into ions.

Can we obtain this chemical energy theoretically by computing how much work it would take to pull apart the crystal? According to our theory, this work is the sum of the potential energies of all the pairs of ions. The easiest way to figure out this sum is to pick out a particular ion and compute its potential energy with each of the other ions. That will give us twice the energy per ion, because the energy belongs to the pairs of charges.

If we want the energy to be associated with one particular ion, we should take half the sum. But we really want the energy per molecule , which contains two ions, so that the sum we compute will give directly the energy per molecule. We are considering monovalent ions.

But it is still a long way from the infinite sum of terms we need. Considering that the ion marked Na in Fig. Now consider the next adjacent line of ions above. Then there are the four lines which are the nearest lines on diagonals, and on and on. It shows that our idea that the whole lattice is held together by electrical Coulomb forces is fundamentally correct. This is the first time that we have obtained a specific property of a macroscopic substance from a knowledge of atomic physics.

We will do much more later. The subject that tries to understand the behavior of bulk matter in terms of the laws of atomic behavior is called solid-state physics. Now what about the error in our calculation? Why is it not exactly right? It is because of the repulsion between the ions at close distances. They are not perfectly rigid spheres, so when they are close together they are partly squashed. They are not very soft, so they squash only a little bit.

Some energy, however, is used in deforming them, and when the ions are pulled apart this energy is released. The actual energy needed to pull the ions apart is a little less than the energy that we calculated; the repulsion helps in overcoming the electrostatic attraction. Is there any way we can make an allowance for this contribution? We could if we knew the law of the repulsive force.

We are not ready to analyze the details of this repulsive mechanism, but we can get some idea of its characteristics from some large-scale measurements. From a measurement of the compressibility of the whole crystal, it is possible to obtain a quantitative idea of the law of repulsion between the ions and therefore of its contribution to the energy.

If a correction is made for this effect, very good agreement with the experimental number is obtained. The ideas are then correct; the major contribution to the energy of a crystal like NaCl is electrostatic. We will now take up another example of electrostatic energy in atomic physics, the electrical energy of atomic nuclei.

Before we do this we will have to discuss some properties of the main forces called nuclear forces that hold the protons and neutrons together in a nucleus. In the early days of the discovery of nuclei—and of the neutrons and protons that make them up—it was hoped that the law of the strong, nonelectrical part of the force between, say, a proton and another proton would have some simple law, like the inverse square law of electricity.

For once one had determined this law of force, and the corresponding ones between a proton and a neutron, and a neutron and a neutron, it would be possible to describe theoretically the complete behavior of these particles in nuclei.

Therefore a big program was started for the study of the scattering of protons, in the hope of finding the law of force between them; but after thirty years of effort, nothing simple has emerged. A considerable knowledge of the force between proton and proton has been accumulated, but we find that the force is as complicated as it can possibly be.

First, the force is not a simple function of the distance between the two protons. At large distances there is an attraction, but at closer distances there is a repulsion. The distance dependence is a complicated function, still imperfectly known. The protons have a spin, and any two interacting protons may be spinning with their angular momenta in the same direction or in opposite directions. And the force is different when the spins are parallel from what it is when they are antiparallel, as in a and b of Fig.

The difference is quite large; it is not a small effect. Third, the force is considerably different when the separation of the two protons is in the direction parallel to their spins, as in c and d of Fig. Fourth, the force depends, as it does in magnetism, on the velocity of the protons, only much more strongly than in magnetism. And this velocity-dependent force is not a relativistic effect; it is strong even at speeds much less than the speed of light.

Furthermore, this part of the force depends on other things besides the magnitude of the velocity. For instance, when a proton is moving near another proton, the force is different when the orbital motion has the same direction of rotation as the spin, as in e of Fig. The force between a proton and a neutron and between a neutron and a neutron are also equally complicated. To this day we do not know the machinery behind these forces—that is to say, any simple way of understanding them.

There is, however, one important way in which the nucleon forces are simpler than they could be. That is that the nuclear force between two neutrons is the same as the force between a proton and a neutron, which is the same as the force between two protons!

If, in any nuclear situation, we replace a proton by a neutron or vice versa , the nuclear interactions are not changed. This fact is nicely illustrated by the locations of the energy levels in similar nuclei. In the nucleus the eleven particles interact with one another in a most complicated dance. Now, there is one configuration of all the possible interactions which has the lowest possible energy; this is the normal state of the nucleus, and is called the ground state.

If the nucleus is disturbed for example, by being struck by a high-energy proton or other particle it can be put into any number of other configurations, called excited states , each of which will have a characteristic energy that is higher than that of the ground state.

The lowest horizontal line represents the ground state. The study of nuclear physics attempts to find an explanation for this rather complicated pattern of energies; there is as yet, however, no complete general theory of such nuclear energy levels.

The broken lines indicate levels for which the experimental information is questionable. Looking at Fig. And so on. After about the tenth level, the correspondence seems to become lost, but can still be seen if the levels are labeled with their other defining characteristics—for instance, their angular momentum and what they do to lose their extra energy.

It must reveal some physical law. It shows, in fact, that even in the complicated situation in a nucleus, replacing a neutron by a proton makes very little change.