Revisited

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For some function f(x) we already know that

(Eq'n 1)

But we can't use that equation as a solution because the integrand of our integral does not give us a proper differential of the function inside the root sign. Nonetheless, we expect that our solution will bear some resemblance to Equation 1, so we postulate that

(Eq'n 2)

in which g(x) represents a to-be-determined function of the variable x.

If we differentiate Equation 2, we get

(Eq'n 3)

Multiplying that equation by the cube of the square root and then rearranging by addition gives us

(Eq'n 4)

in which g'(x) = dg/dx. We divide that equation by dx, make an educated guess, and split it into two equations;

(Eq'n 5)

and

(Eq'n 6)

which give us g = x/a2, so we have Equation 2, our solution, as

(Eq'n 7)

Now I want to solve that integral by a method that requires no educated guesses, but does require a knowledge of infinite series. I begin by factoring the constant out of the square root and making the substitution x=ay, so I get

(Eq'n 8)

I then replace the integrand by its equivalent infinite series and get

(Eq'n 9)

Term by term integration of that series representation gives us

(Eq'n 10)

Next I factor y out of the series and recognize the series as a root of the alternating geometric series to get

(Eq'n 11)

Finally I make the substitution y=x/a and pull one factor of a into the square root to obtain

(Eq'n 12)

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Example Application:

Electric Field of an Infinitely Long Wire

We want to calculate the strength of an electric field emanating from a uniformly charged, infinitely-long, straight wire that lies on the y-axis of our coordinate grid. We describe the charge on the wire as having a linear density (Coulombs per meter) of λ. And we invoke symmetry to infer that at any field point at a point (X,0,0) on the x-axis the electric field points only in the x-direction. Now we have two ways in which we can determine the strength of the electric field at the field point.

1) We can finesse the calculation through Gauss's law; We select a segment of the wire, of length dy, passing through the origin of our coordinate frame and enclose it in a right circular cylinder of altitude dy and radius X. We so position that cylinder that our selected segment of wire lies entirely on its axis of circular symmetry. In that configuration the faces of the cylinder contribute nothing to the electric flux through the cylinder's surface. On the side of the cylinder the electric field has the same strength everywhere (again by symmetry) and points in the direction perpendicular to the surface, so for the electric flux we have

(Eq'n 13)

The latter equality in that equation comes from Gauss's law, the first of Maxwell's Equations, which relates the electric flux through a closed surface to the amount of electric charge enclosed within that surface. Solving that equation for the electric field strength gives us directly

(Eq'n 14)

2) We can force the calculation through Coulomb's law; A segment of the wire, of length dy, a distance y from the origin of our coordinate grid generates an electric field of strength

(Eq'n 15)

at the field point. Of that field we want only the component that points in the x-direction (the y-ward components all cancel out), so we apply the standard right triangle of vector addition and the Pythagorean theorem to calculate the proportion of the field strength in Equation 15 that points in the x-direction:

(Eq'n 16)

To calculate the total field strength at the field point we must add up that contribution as it comes from each and every element of electric charge on the wire and we accomplish that calculation by integrating Equation 16 over the domain extending from y=-∞ to y=+∞:

(Eq'n 17)

We cannot properly carry out a calculation with infinities, so we must employ the limiting process. We notice that as y becomes extremely large relative to X we can make the approximation

(Eq'n 18)

As the value of y increases, tending toward infinity, that approximation comes proportionately closer to the actual value of the square root, so we have the limit

(Eq'n 19)

We apply that fact to Equation 17 and note that subtracting a negative number from a positive number of the same magnitude simply doubles the positive number and obtain

(Eq'n 20)

which is what we got from applying Gauss's law.

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