. Gauss's Law

We already know about electric field lines and electricflux. Electricfluxthrough a closed surface S is

which is the number of field lines passing through surface S.

  • Statement of Gauss's Law
    “Electricfluxthrough any surface enclosing charge is equal to q/ε0, where q is the net charge enclosed by the surface”
    mathematically,

    where qencis the net charge enclosed by the surface andEis the total electric field at each point on the surface under consideration.
  • It is the net charge enclosed in the surface that matters in Gauss's law but the totalfluxof electric fieldEdependsalso on the surface chosen not merely on the charge enclosed.
  • So if you have information about distribution ofelectric chargeinside the surface you can find electricfluxthrough that surface using Gauss's Law.
  • Again if you have information regarding electric fluxthrough any closed surface then total charge enclosed by that surface can also be easily calculated using Gauss's Law.
  • Surface on which Gauss's Law is applied is known as Gaussian surface which need not be a real surface.
  • Gaussian surface can be an imaginary geometrical surface which might be empty space or it could be partially or fully embedded in a solid body.
  • Again consider equation 11

    In left hand side of above equationE·dais scalar product of two vectors namely electric field vectorEand area vectorda. Area vectordais defined as the vector of magnitude |da| whose direction is that of outward normal to area element da. So,da=nˆda wherenˆ is unit vector along outward normal to da.


    From above discussion we can conclude that,
    (1)If bothEand surface area da at each points are perpendicular to each other and has same magnitude at all points of the surface then vectorEhas same direction as that of area vector as shown below in the figure.

    sinceEis perpendicular to the surface

    (2) IfEis parallel to the surface as shown below in the figure


    E at all points on the surface.

APPLICATIONS OF GAUSS'S LAW

* Derivation of Coulomb’s Law.

  • Coulomb’s law can be derived from Gauss's law.
  • Consider electric field of a single isolated positive charge of magnitude q as shown below in the figure.

  • Field of a positive charge is in radially outward direction everywhere and magnitude of electric field intensity is same for all points at a distance r from the charge.
  • We can assume Gaussian surface to be a sphere of radius r enclosing the charge q.
  • From Gauss's law

    sinceEis constant at all points on the surface therefore,

    surface area of the sphere is A=4πr2
    thus,

    Now force acting on point charge q' at distance r from point charge q is

    This is mathematical statement of Coulomb’s law.

* Electric field due to line charge

Consider a long thin uniformly charged wire and we have to find the electric field intensity due to the wire at any point at perpendicular distance from the wire.

If the wire is very long and we are at point far away from both its ends then field lines outside the wire are radial and would lie on a plane perpendicular to the wire.

Electric field intensity has same magnitude at all points which are at same distance from the line charge.

We can assume Gaussian surface to be a right circular cylinder of radius r and length l with its ends perpendicular to the wire as shown below in the figure.

λ is the charge per unit length on the wire. Direction ofEis perpandicular to the wire and components ofEnormal to end faces of cylinder makes no contribution to electric flux. Thus from Gauss's law

Now consider left hand side of Gauss's lawsince at all points on the curved surfaceEis constant. Surface area of cylinder of radius r and length l is A=2πrl therefore,

Charge enclosed in cylinder is q=linear charge density x length l of cylinder,or, q=λlFrom Gauss's lawThus electric field intensity of a long positively charged wire does not depends on length of the wire but on the radial distance r of points from the wire.

Electric field due to charged solid sphere

  • We'll now apply Gauss's law to find the field outside uniformly charged solid sphere of radius R and total charge q.
  • In this case Gaussian surface would be a sphere of radius r>Rconcentricwith the charged solid sphere shown below in the figure.From Gauss's law

    where q is the charge enclosed.
  • Charge is distributed uniformly over the surface of the sphere. Symmetry allows us to extractEout of theintegralsign asmagnitudeof electric fieldintensityis same for all points at distance r>R.
  • Since electric field points radially outwards we have


    also as discussedmagnitudeofEis constant over Gaussian surface so,

    where 4πr2is the surface area of the sphere.
    Again from Gauss's law we have

    Thus we see thatmagnitudeof field outside the sphere is exactly the same as it would have been as if all the charge were concentrated at its center.

* Electric field due to an infinite plane sheet of charge

  • Consider a thin infinite plane sheet of charge having surface charge density σ(charge per unit area).
  • We have to find the electric fieldintensitydue to this sheet at any point which is distance r away from the sheet.
  • We can draw a rectangular Gaussian pillbox extending equal distance above and below the plane as shown below in the figure.

  • By symmetry we find thatEon either side of sheet must be perpandicular to the plane of the sheet, having samemagnitudeat all points equidistant from the sheet.
  • No field lines crosses the side walls of the Gaussian pillbox i.e., component ofEnormal to walls of pillbox is zero.
  • We now apply Gauss's law to this surface

    in this case charge enclosed is
    q=σA
    where A is the area of end face of Gaussian pillbox.
  • Epoints in the direction away from the plane i.e.,Epoints upwards for points above the plane anddownwardsfor points below the plane. Thus for top and bottom surfaces,

    thus
    2A|E|=σA/ε0
    or,
    |E|=σ/2ε0
    Here one important thing to note is thatmagnitudeof electric field at any point is independent of the sheet and does not decrease inversely with the square of the distance. Thus electric field due to an infinite plane sheet of charge does not falls of at all.

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Electric Potential and Electric Field

Consider a chargeplaced in an electric field generated by fixed charges. Let us chose some arbitrary reference pointin the field. At this point, the electric potential energy of the charge is defined to be zero. Thisuniquelyspecifies the electric potential energy of the charge at every other point in the field. For instance, the electric potential energyat some pointis simply the workdone in moving the charge fromtoalong any path. As, we can calculate. It is clear, from (from previous knowledge of , thatdepends both on the particular chargewhich we place in the field, and the magnitude and direction of the electric field along the chosen route between pointsand. However, it is also clear thatisdirectly proportionalto the magnitude of the charge. Thus, if the electric potential energy of a chargeat pointisthen the electric potential energy of a chargeat the same point is. We can exploit this fact to define a quantity known as theelectric potential. The difference in electric potential between two pointsandin an electric field is simply the work done in moving some charge between the two points divided by the magnitude of the charge. Thus,

/ (80)

wheredenotes the electric potential at point,etc.This definition uniquely defines the difference in electric potential between pointsand, but the absolute value of the potential at pointremains arbitrary. We can therefore, without loss of generality, set the potential at pointequal to zero. It follows that the potential energy of a chargeat some pointis simply the product of the magnitude of the charge and the electric potentialat that point:

/ (81)

The electric potential at point(relative to point) is solely a property of the electric field, and is, therefore, the same for any charge placed at that point. We shall see exactly how the electric potential is related to the electric field later on.

The dimensions of electric potential are work (or energy) per unit charge. The units of electric potential are, therefore, joules per coulomb (). A joule per coulomb is usually referred to as a volt (V):i.e.,

/ (82)

Thus, the alternative (and more conventional) units of electric potential are volts. The difference in electric potential between two points in an electric field is usually referred to as thepotential difference, or even the difference in ``voltage,'' between the two points.

A battery is a convenient tool for generating a difference in electric potential between two points in space. For instance, a twelve volt (12V) battery generates an electric field, usually via some chemical process, which is such that the potential differencebetween its positive and negative terminals is twelve volts. This means that in order to move a positive charge of 1 coulomb from the negative to the positive terminal of the battery we must do 12 joules of work against the electric field. (This is true irrespective of the route taken between the two terminals). This implies that the electric field must be directed predominately from the positive to the negative terminal.

More generally, in order to move a chargethrough a potential differencewe must do work, and the electric potential energy of the charge increases by an amountin the process. Thus, if we move an electron, for whichC, through a potential difference of minus 1 volt then we must dojoules of work. This amount of work (or energy) is called anelectron volt(eV):i.e.,

/ (83)

The electronvolt is a convenient measure of energy in atomic physics. For instance, the energy required to break up a hydrogen atom into a free electron and a free proton iseV.

We have seen that the difference in electric potential between two arbitrary points in space is a function of the electric field which permeates space, but is independent of the test charge used to measure this difference. Let us investigate the relationship between electric potential and the electric field.

Consider a chargewhich is slowly moved an infinitesimal distancealong the-axis. Suppose that the difference in electric potential between the final and initial positions of the charge is. By definition, the changein the charge's electric potential energy is given by

/ (84)

From Eq.(76), the workwhich we perform in moving the charge is

/ (85)

whereis the local electric field-strength, andis the angle subtended between the direction of the field and the-axis. By definition,, whereis the-component of the local electric field. Energy conservation demands that(i.e., the increase in the charge's energy matches the work done on the charge), or

/ (86)

which reduces to

/ (87)

We call the quantitythegradientof the electric potential in the-direction. It basically measures how fast the potentialvaries as the coordinateis changed (but the coordinatesandare held constant). Thus, the above formula is saying that the-component of the electric field at a given point in space is equal tominusthe local gradient of the electric potential in the-direction.

According to Eq.(87), electric field strength has dimensions of potential difference over length. It follows that the units of electric field are volts per meter (. Of course, these new units are entirely equivalent to newtons per coulomb:i.e.,

/ (88)

Consider the special case of a uniform-directed electric fieldgenerated by two uniformly charged parallel planes normal to the-axis. It is clear, from Eq.(87), that ifis to be constant between the plates thenmust varylinearlywithin this region. In fact, it is easily shown that

/ (89)

whereis an arbitrary constant. According to Eq.(89), the electric potentialdecreasescontinuously as we move along the direction of the electric field. Since a positive charge is accelerated in this direction, we conclude that positive charges are accelerateddowngradients in the electric potential, in much the same manner as masses fall down gradients of gravitational potential (which is, of course, proportional to height). Likewise, negative charges are acceleratedupgradients in the electric potential.

According to Eq.(87), the-component of the electric field is equal to minus the gradient of the electric potential in the-direction. Since there is nothing special about the-direction, analogous rules must exist for the- and-components of the field. These three rules can be combined to give

/ (90)

Here, thederivative is taken at constantand,etc.The above expression shows how the electric field, which is a vector field, is related to the electric potential, which is a scalar field.

We have seen that electric fields are superposable. That is, the electric field generated by a set of charges distributed in space is simply thevector sumof the electric fields generated by each charge taken separately. Well, if electric fields are superposable, it follows from Eq.(90) that electric potentials must also be superposable. Thus, the electric potential generated by a set of charges distributed in space is just thescalar sumof the potentials generated by each charge taken in isolation. Clearly, it is far easier to determine the potential generated by a set of charges than it is to determine the electric field, since we can sum the potentials generated by the individual charges algebraically, and do not have to worry about their directions (since they have no directions).

Equation(90) looks rather forbidding. Fortunately, however, it is possible to rewrite this equation in a more appealing form. Consider two neighboring pointsand. Suppose thatis the vector displacement of pointrelative to point. Letbe the difference in electric potential between these two points. Suppose that we travel fromtoby first moving a distancealong the-axis, then movingalong the-axis, and finally movingalong the-axis. The net increase in the electric potentialas we move fromtois simply the sum of the increasesas we move along the-axis,as we move along the-axis, andas we move along the-axis:

/ (91)

But, according to Eq.(90),,etc.So, we obtain

/ (92)

which is equivalent to

/ (93)

whereis the angle subtended between the vectorand the local electric field. Note thatattains its most negative value when. In other words, the direction of the electric field at pointcorresponds to the direction in which the electric potentialdecreases most rapidly. A positive charge placed at pointis accelerated in this direction. Likewise, a negative charge placed atis accelerated in the direction in which the potential increases most rapidly (i.e.,). Suppose that we move from pointto a neighboring pointin a direction perpendicular to that of the local electric field (i.e.,). In this case, it follows from Eq.(93) that the pointsandlie at the same electric potential (i.e.,). The locus of all the points in the vicinity of pointwhich lie at the same potential asis a plane perpendicular to the direction of the local electric field. More generally, the surfaces of constant electric potential, the so-calledequipotential surfaces, exist as a set of non-interlocking surfaces which are everywhere perpendicular to the direction of the electric field. Figure14shows the equipotential surfaces (dashed lines) and electric field-lines (solid lines) generated by a positive point charge. In this case, the equipotential surfaces are spheres centred on the charge.

Figure 14:The equipotential surfaces (dashed lines) and the electric field-lines (solid lines) of a positive point charge.

In Sect.4.3, we found that the electric field immediately above the surface of a conductor is directed perpendicular to that surface. Thus, it is clear that the surface of a conductor must correspond to an equipotential surface. In fact, since there is no electric field inside a conductor (and, hence, no gradient in the electric potential), it follows that the whole conductor (i.e., both the surface and the interior) lies at the same electric potential.

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MULTIPOLE EXPANSION

A multipole expansion is a series expansion of the effect produced by a given system in terms of an expansion parameter which becomes small as the distance away from the system increases. Therefore, the leading one or terms in a multipole expansion are generally the strongest. The first-order behavior of the system at large distances can therefore be obtained from the first terms of this series, which is generally much easier to compute than the general solution. Multipole expansions are most commonly used in problems involving the gravitational field of mass aggregations, the electric and magnetic fields of charge and current distributions, and the propagation of electromagnetic waves.

Electric Multipole Expansion

For a given charge distribution, we can write down amultipole expansion, which gives the potential as a series in powers of, whereis the distance from the origin to the observation point.

We know that thepotential in generalis

In the integral,is the position of charge element. From the law of cosines

whereis the angle betweenand. We can rewrite this as

From the theory of Legendre polynomials, it is known that the last factor in this expression is agenerating functionfor the polynomials. That is, if we write the square root as an power series, we get

The coefficient ofin the series is the Legendre polynomial. This can be verified for the first few terms by calculating the Taylor series expansion of the square root term about. This is tedious to do by hand, but using Maple, we get, defining: