The electric field has already been described in terms of the force on a charge. If the electric potential is known at every point in a region of space, the electric field can be derived from the potential. In vector calculus notation, the electric field is given by the negative of the gradient of the electric potential, E = −grad V. This expression specifies how the electric field is calculated at a given point. Since the field is a vector, it has both a direction and magnitude. The direction is that in which the potential decreases most rapidly, moving away from the point. The magnitude of the field is the change in potential across a small distance in the indicated direction divided by that distance.

To become more familiar with the electric potential, a numerically determined solution is presented for a two-dimensional configuration of electrodes. A long, circular conducting rod is maintained at an electric potential of −20 volts. Next to the rod, a long L-shaped bracket, also made of conducting material, is maintained at a potential of +20 volts. Both the rod and bracket are placed inside a long, hollow metal tube with a square cross section; this enclosure is at a potential of zero (i.e., it is at “ground” potential). Figure 6 shows the geometry of the problem. Because the situation is static, there is no electric field inside the material of the conductors. If there were such a field, the charges that are free to move in a conducting material would do so until equilibrium was reached. The charges are arranged so that their individual contributions to the electric field at points inside the conducting material add up to zero. In a situation of static equilibrium, excess charges are located on the surface of conductors. Because there are no electric fields inside the conducting material, all parts of a given conductor are at the same potential; hence, a conductor is an equipotential in a static situation.

In Figure 7, the numerical solution of the problem gives the potential at a large number of points inside the cavity. The locations of the +20-volt and −20-volt electrodes can be recognized easily. In carrying out the numerical solution of the electrostatic problem in the figure, the electrostatic potential was determined directly by means of one of its important properties: in a region where there is no charge (in this case, between the conductors), the value of the potential at a given point is the average of the values of the potential in the neighbourhood of the point. This follows from the fact that the electrostatic potential in a charge-free region obeys Laplace’s equation, which in vector calculus notation is div grad V = 0. This equation is a special case of Poisson’s equation div grad V = ρ, which is applicable to electrostatic problems in regions where the volume charge density is ρ. Laplace’s equation states that the divergence of the gradient of the potential is zero in regions of space with no charge. In the example of Figure 7, the potential on the conductors remains constant. Arbitrary values of potential are initially assigned elsewhere inside the cavity. To obtain a solution, a computer replaces the potential at each coordinate point that is not on a conductor by the average of the values of the potential around that point; it scans the entire set of points many times until the values of the potentials differ by an amount small enough to indicate a satisfactory solution. Clearly, the larger the number of points, the more accurate the solution will be. The computation time as well as the computer memory size requirement increase rapidly, however, especially in three-dimensional problems with complex geometry. This method of solution is called the “relaxation” method.

In Figure 8, points with the same value of electric potential have been connected to reveal a number of important properties associated with conductors in static situations. The lines in the figure represent equipotential surfaces. The distance between two equipotential surfaces tells how rapidly the potential changes, with the smallest distances corresponding to the location of the greatest rate of change and thus to the largest values of the electric field. Looking at the +20-volt and +15-volt equipotential surfaces, one observes immediately that they are closest to each other at the sharp external corners of the right-angle conductor. This shows that the strongest electric fields on the surface of a charged conductor are found on the sharpest external parts of the conductor; electrical breakdowns are most likely to occur there. It also should be noted that the electric field is weakest in the inside corners, both on the inside corner of the right-angle piece and on the inside corners of the square enclosure.

In Figure 9, dashed lines indicate the direction of the electric field. The strength of the field is reflected by the density of these dashed lines. Again, it can be seen that the field is strongest on outside corners of the charged L-shaped conductor; the largest surface charge density must occur at those locations. The field is weakest in the inside corners. The signs of the charges on the conducting surfaces can be deduced from the fact that electric fields point away from positive charges and toward negative charges. The magnitude of the surface charge density σ on the conductors is measured in coulombs per metre squared and is given bywhere ε₀ is called the permittivity of free space and has the value of 8.854 × 10⁻¹² coulomb squared per newton-square metre. In addition, ε₀ is related to the constant k in Coulomb’s law by

Britannica Quiz

Electricity: Short Circuits & Direct Currents

Figure 9 also illustrates an important property of an electric field in static situations: field lines are always perpendicular to equipotential surfaces. The field lines meet the surfaces of the conductors at right angles, since these surfaces also are equipotentials. Figure 10 completes this example by showing the potential energy landscape of a small positive charge q in the region. From the variation in potential energy, it is easy to picture how electric forces tend to drive the positive charge q from higher to lower potential—i.e., from the L-shaped bracket at +20 volts toward the square-shaped enclosure at ground (0 volts) or toward the cylindrical rod maintained at a potential of −20 volts. It also graphically displays the strength of force near the sharp corners of conducting electrodes.

Principle of the capacitor

To understand how a charged capacitor stores energy, consider the following charging process. With both plates of the capacitor initially uncharged, a small amount of negative charge is removed from the lower plate and placed on the upper plate. Thus, little work is required to make the lower plate slightly positive and the upper plate slightly negative. As the process is repeated, however, it becomes increasingly difficult to transport the same amount of negative charge, since the charge is being moved toward a plate that is already negatively charged and away from a plate that is positively charged. The negative charge on the upper plate repels the negative charge moving toward it, and the positive charge on the lower plate exerts an attractive force on the negative charge being moved away. Therefore, work has to be done to charge the capacitor.

Where and how is this energy stored? The negative charges on the upper plate are attracted toward the positive charges on the lower plate and could do work if they could leave the plate. Because they cannot leave the plate, however, the energy is stored. A mechanical analogy is the potential energy of a stretched spring. Another way to understand the energy stored in a capacitor is to compare an uncharged capacitor with a charged capacitor. In the uncharged capacitor, there is no electric field between the plates; in the charged capacitor, because of the positive and negative charges on the inside surfaces of the plates, there is an electric field between the plates with the field lines pointing from the positively charged plate to the negatively charged one. The energy stored is the energy that was required to establish the field. In the simple geometry of Figure 11, it is apparent that there is a nearly uniform electric field between the plates; the field becomes more uniform as the distance between the plates decreases and the area of the plates increases. It was explained above how the magnitude of the electric field can be obtained from the electric potential. In summary, the electric field is the change in the potential across a small distance in a direction perpendicular to an equipotential surface divided by that small distance. In Figure 11, the upper plate is assumed to be at a potential of V_a volts, and the lower plate at a potential of V_b volts. The size of the electric field isin volts per metre, where d is the separation of the plates. If the charged capacitor has a total charge of +Q on the inside surface of the lower plate (it is on the inside surface because it is attracted to the negative charges on the upper plate), the positive charge will be uniformly distributed on the surface with the valuein coulombs per metre squared. Equation (8) gives the electric field when the surface charge density is known as E = σ/ε₀. This, in turn, relates the potential difference to the charge on the capacitor and the geometry of the plates. The result is

The quantity C is termed capacity; for the parallel-plate capacitor, C is equal to ε₀A/d. The unit used for capacity is the farad (F); one farad equals one coulomb per volt. In equation (12), only the potential difference is involved. The potential of either plate can be set arbitrarily without altering the electric field between the plates. Often one of the plates is grounded—i.e., its potential is set at the Earth potential, which is referred to as zero volts. The potential difference is then denoted as ΔV, or simply as V.

Three equivalent formulas for the total energy W of a capacitor with charge Q and potential difference V are

All are expressed in joules. The stored energy in the parallel-plate capacitor also can be expressed in terms of the electric field; it is, in joules,

The quantity Ad, the area of each plate times the separation of the two plates, is the volume between the plates. Thus, the energy per unit volume (i.e., the energy density of the electric field) is given by 1/2ε₀E² in units of joules per metre cubed.