DSC1BT (Electricity and Magnetism) Topic – Electrostatics (Part – 2)
We have already discussed part 1 of this e-report.
Now let us continue part 2 of it.
Electric Field and Flux due to Point Charge:
Let us take this simplest example. Suppose that the field is that of a single isolated positive point charge , and the surface is a sphere of radius centred on the point charge (shown in Fig. 1).
Fig. 1
Now we are interested in the flux through this surface. The answer is simple to calculate because the magnitude of at every point on the surface is given by ( )
and its direction is the same as that of the outward normal at that point. So we have
Therefore, we see that the flux is independent of the size or radius of the sphere.
Here for the first time we see the benefit of including the factor of
in Coulomb’s Law. Without this factor, we would have a retained factor of in the previous equation and therefore also, eventually, in one of Maxwell’s Equations.
Fig. 2
Now let us imagine a second surface enclosing the first, but not spherical in shape, as in Fig. 2. We will find that the total flux through this arbitrary surface is the same as that through the sphere. To see this, let us concentrate on a cone, radiating from , that cuts a small patch out of the sphere and continues on to the outer surface, where it cuts out a patch at a distance from the point charge. The area of the patch is larger than that of the patch by two factors.
Firstly, by the ratio of the distance squared and secondly, owing to its inclination, by the factor
. The angle is the angle between the outward normal and the radial direction (see Fig. 2). The electric field in that neighbour- hood is reduced from its magnitude on the sphere by the factor and is still radially directed. Letting be the field at the outer patch and be the field at the sphere, we have the flux through outer patch
and the flux through the inner patch
.
Using the above facts concerning the magnitude of and the area of , the flux through the outer patch can be written as
which equals the flux through the inner patch .
Now every patch on the outer surface can in this way be put into correspondence with part of the spherical surface, so the total flux must be the same through the two surfaces. That is, the flux through the new surface must be just . But this was a surface of arbitrary shape and size. So, we conclude that the flux of the electric field through any surface enclosing a point charge is . As a corollary we can say that the total flux through a closed surface is zero if the charge lies outside the surface, which is basically the statement of Gauss’s Theorem.
Gauss’s Theorem:
Any electric field is the sum of the fields of its individual sources. This property was expressed in the statement of Coulomb’s Law. Clearly flux is an additive quantity in the same sense, for if we have a number of sources, , , . . . ,
the fields of which, if each were present alone, would be , , . . . , , then the flux through some surface in the actual field can be written
We have learnt that equals to if the th charge is enclosed by and equals zero otherwise. So every charge inside the surface contributes exactly to the surface integral and all charges outside contribute nothing. This is nothing but Gauss Theorem, which states that the flux of the electric field through any closed surface, that is, the integral over the surface, equals
times the total charge enclosed by the surface, say .
The total enclosed charge can be written in terms of volume integral of volume charge density as . Then Gauss’s Theorem becomes
.
This remarkable theorem extends our knowledge in two ways. First, it reveals a connection between the field and its sources that is the converse of Coulomb’s Law. Coulomb’s Law tells us how to derive the electric field if the charges are given, but with Gauss’s Theorem we can determine how much charge is in any region if the field is known. Second, the mathematical relation here demonstrated is a powerful analytic tool, it can make complicated problems easy, as we shall see in the following few examples.
Electric Field due to Spherical Charge Distribution:
We can use Gauss’s Theorem to find the electric field of a spherically symmetrical distribution of charge, that is, a distribution in which the volume charge density depends only on the radius from a central point. Fig. 3 depicts a cross section through some such distribution. Here the charge density is high at the centre, and is zero beyond a certain boundary ( ).
Because of the spherical symmetry, the electric field at any point must be radially directed, no other direction is unique. Likewise, the field magnitude must be the same at all points on a spherical surface of radius , for all such points are equivalent. We call this field magnitude as . The flux through this surface is therefore , and by Gauss’s Theorem this must be equal to
times the charge enclosed by the surface ( ) . Therefore, or
.
Fig. 3
Comparing this with the field of a point charge, we see that the field at all points on is the same as if all the charge within were concentrated at the centre.
The same statement applies to a sphere drawn inside the charge distribution.
The field at any point on is the same as if all charge within were at the centre and all charge outside absent. Evidently the field inside a hollow spherical charge distribution is zero ( ).
Field Outside Uniformly Charged Sphere. Here the volume charge density is constant upto a distance from the centre. For an external point , the field is the same as if all of the charge were concentrated at the centre of the
sphere. Since the volume of the sphere is , the field is therefore radial and has magnitude given by
The electric field therefore falls off as outside the sphere and eventually becomes zero as .
Field Inside Uniformly Charged Sphere. Here for , the charge outside radius effectively contributes nothing to the field, while the charge inside radius acts as if it were concentrated at the centre. The volume inside radius is , so the field inside the given sphere is radial and has magnitude
In terms of the total charge , the electric field can be written as
. The field increases linearly with inside the sphere since the growth of the effective charge outweighs the effect from the increasing distance. A plot of for the entire range of is shown in Fig. 4. It is noteworthy that is continuous at , where it takes on the value .
Fig. 4
Electric Field due to Infinite Line Charge Distribution:
An infinitely long, straight, charged wire, if we neglect its thickness, can be characterized by the amount of charge it carries per unit length. Let denotes its linear charge density.
We will do the problem in two ways, first by an integration starting from Coulomb’s Law, and then by using Gauss’s Theorem.
Fig. 5
To evaluate the field at the point P, shown in Fig. 5, we must add up the contributions from all segments of the line charge, one of which is indicated as a segment of length . The charge on this element is given by . Having oriented the -axis along the line charge, we may as well let the -axis pass through P, which is a distance from the nearest point on the line. It is a good idea to take advantage of symmetry at the outset. Obviously the electric field at P must point in the direction, so that and are both zero. The contribution of the charge to the component of the electric field at P is given by
where is the angle the electric field of makes with the direction. The total component is then
.
Using the relation between and as , we get . Changing the variable of integration from to with proper limits, we write
.
We see that the field of an infinitely long, uniformly dense line charge is proportional to the reciprocal of the distance from the line. Its direction is of course radially outward if the line carries a positive charge, inward if negative.
Fig. 6
Gauss’s Theorem leads directly to the same result. In order to do that, let us surround a segment of the line charge with a closed circular cylinder (or tin can) of length and radius , as shown in Fig. 6 and consider the flux through this surface. As we have already noted, symmetry guarantees that the field is radial, so the flux through the ends of the tin can is zero. The flux through the cylindrical surface is simply the area, times , the field at the surface. On the other hand, the charge enclosed by the surface is just , so Gauss’s Theorem gives us or
which matches with the previous result.
Electric Field due to Infinite Thin Sheet of Charge:
Fig. 7
Electric charge distributed smoothly in a thin sheet is called a surface charge distribution. Let us consider a flat sheet, infinite in extent, with the constant surface charge density . The electric field on either side of the sheet, whatever its magnitude may turn out to be, must surely point perpendicular to the plane of the sheet, so there is no other unique direction in the system. Also because of symmetry, the field must have the same magnitude and the opposite direction at two points P and Pʹ equidistant from the sheet on opposite sides. With these facts established, Gauss’s Theorem gives us at once the field intensity, as follows. Let us draw a cylinder, as in Fig. 7 (actually, any shape with uniform cross section will work fine), with P on one side and Pʹ on the other, of cross-
sectional area . The outward flux is found only at the ends, so that if denotes the magnitude of the field at P, and the magnitude at Pʹ, the outward flux is given by . The charge enclosed is and therefore according to Gauss’s Theorem or
.
We see that the field strength is independent of , the distance from the sheet.
The previous equation could have been derived more laboriously by calculating the vector sum of the contributions to the field at P from all the little elements of charge in the sheet.
In a more general case where there are other charges in the vicinity, the field need not be perpendicular to the sheet, or might not be symmetric on either side of it. In that situation, let us consider a very squat Gaussian surface or pillbox, with P and Pʹ infinitesimally close to the sheet, instead of the elongated surface in Fig. 7. We can then ignore the negligible flux through the cylindrical side of the pillbox, so the above reasoning gives
where the sign denotes the component perpendicular to the sheet. If one wants to write this in terms of vectors, it becomes , where is the unit vector perpendicular to the sheet, in the direction of P. In other words, the discontinuity in across the sheet is given by
Only the normal component is discontinuous and the parallel component is continuous across the sheet. So we can just replace the in the previous equation with . This result is also valid for any finite-sized sheet, because from up close the sheet looks essentially like an infinite plane, at least as far as the normal component is concerned.
This concludes part 2 of this e-report.
The discussion will be continuing in the part 3 of this e-report.
Reference(s):
Electricity and Magnetism, E.M. Purcell and D.J. Morin, Cambridge University Press
Electromagnetism: Theory and Applications, Ashutosh Pramanik, PHI (All the figures have been collected from the above mentioned references)
