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Example 1- Electric field of a point charge
All right. As a first application of the Gauss’s law, let’s try to calculate the electric field of a point charge, which we already know from Coulomb’s law, the electric field of a point charge. Let’s see how we can get the same result by applying Gauss’s law. Let’s assume that we have a positive point charge sitting over here and it generates its own electric field in radially outward direction originating from the charge and going to infinity, filling the whole space surrounding the charge.
We’re going to choose our hypothetical surface, S, in the form of a spherical surface. We will choose that such that the side surface of the sphere will pass through our point of interest. If we’re interested with the electric field that this charge generates at this point, therefore we will choose a spherical surface such that it passes through that point of interest. I’m going to use dashed line for this hypothetical surface that we will choose to apply Gauss’s law and this surface is also known as “Gaussian surface”. Since it is hypothetical that we choose, we’re going to use dashed line in order to avoid confusion with other surfaces in the problems that we are dealing with.
Therefore, if this point of interest is some r distance away from the charge, then the radius of this Gaussian surface we choose will also be equal to little r. At the location of our point of interest, the electric field vector is radially out, generated from this positive charge. If we just visualize a tiny little incremental surface on the surface of this sphere, the area vector of such a surface will be perpendicular to the surface and that too will be pointing in radially outward direction.
Here let me make a point. When we draw the area vector for closed surfaces like the surface of a sphere, the area vector should always be pointing outside of the surface. In other words, it cannot point towards the inside of the surface. This is true for all closed surfaces. If we are dealing with open surfaces like the surface of a rectangle for example, which does not enclose any volume, then any direction, either up or down, will be okay when we draw the surface area of such an open surface.
We will have some restrictions later on when we look at the surfaces defined by current loops. In those cases, we will have a restriction associated with the surface area of that region. For now, since we are dealing with the closed surfaces, the surface area will always point outward from the closed surface, a surface which encloses a volume.
Okay. Now along the surface, if we just go ahead and look at a different point, somewhere over here for example, there also electric field is going to be radially out and the incremental surface element at that location will also have area vector perpendicular to that, pointing in the radially outward direction. As a matter of fact, if we do this throughout the whole region along this surface, we will see that E will be radially out and dA is going to be perpendicular to the surface. Therefore it will also be in the same direction with the electric field vector.
Therefore, we see that such a surface will satisfy the conditions because our first condition was the magnitude of the electric field will be constant everywhere on that surface. Since as long as we’re on the surface of this sphere, we will be the same distance away from the source, therefore the magnitude of the electric field that the source generates in that region or along that region will be constant. The first condition is satisfied.
The second condition was the angle between E and dA, that that should remain constant all the time and we have that situation. We can see wherever we go along this surface, the angle between the electric field vector and the incremental surface vector dA is 0 degrees. Therefore cosine of α is equal to 1.
Now, by writing down the expression for Gauss’s law, which is E dot dA, is equal to q-enclosed over ε0, net charge inside of the region surrounded by the Gaussian surface, divided by 0. The left-hand side, in explicit form will be E magnitude dA magnitude times cosine of 0 integrated over closed surface S is equal to q-enclosed over ε0. Since electric field is constant over this surface, we can take it outside of the integral. The cosine of 0 is nothing but 1.
Moving on for the left-hand side we have then E integral of dA integrated over this closed surface S, is equal to q-enclosed over ε0. Integral of dA over this closed surface S means that we are adding all these tiny, little incremental surfaces on the surface of the sphere to one another along the whole surface. If we do that of course, at the end we are going to end up with the total surface area of this sphere.
Therefore this integral is going to give us nothing but the surface area of the Gaussian sphere. The surface area of a sphere is 4 π times its radius squared. Therefore on the left-hand side we will have E times 4 π r2 is equal to, on the right hand side, q-enclosed over ε0. q-enclosed, by definition, is the net charge inside of the region surrounded by Gaussian sphere or Gaussian surface. That is this whole region. When we look at that region, we see only one charge and that is our source charge q. For this case, therefore, q-enclosed is equal to the total charge q.
Our expression then becomes E is equal to, q-enclosed is q, and if we divide both sides by 4 π r2, we will have 4 π ε0 r2 on the right-hand side. This is the expected result that we obtained earlier from Coulomb’s law such that the electric field of a point charge is equal to, the magnitude is equal to q over 4 π ε0 r2.
Again, one can write this down in vector form if we introduce a unit vector in radial direction, since this electric field is in radial direction, we multiply this by the unit vector r̂ in radial direction, where r̂ is the unit vector in radial direction.