Note: see image q=20uC or `(20*10^-6C)` By following the steps below, you can use an integral to find a more exact value of the potential....

The first expression can be derived by using the definition of the potential of the point charge q at a given point distance r away from the charge:

`V = kq/r`

The charged ring shown on the image can be considered a collection of point, or differential, charges dq. The point P is located on the central axis of the ring, so it is the same distance away from any point on the ring: `sqrt(x^2 + a^2)` . The potential at point P due to the charge dq is then

`dV = k(dq)/sqrt(x^2 + a^2)`

The total potential, by superposition principle, is the sum of the potential due to each differential charge, or the integral:

`V = int dV = int k(dq)/sqrt(x^2 + a^2) = kq_(t)/sqrt(x^2 + y^2)` . `d q_t = int dq ` is the total charge.)

The second part of the question requires to write the expression for the potential difference using the relationship between the potential and electric field:

`DeltaV_(ab) = -int _a ^b vecE*vecds` : the potential difference between the two points equal the negative line integral of the electric field on a line connecting these two points. Since electric field is conservative, this integral is independent of the line chosen.

For the finite charge distribution such as the charged ring on the image, the potential at infinity is zero so one of the points chosen to be at infinity. Then,

`V(x) = -int _oo ^x vecE*dvecs` .

If we choose the line connecting infinity to point P to be along the central axis of the ring, then `vecE*dvecs` is the horizontal component of the electric field.

The magnitude of the electric field due to differential charge dq at point P is

`dE = k(dq)/(x^2 + r^2)`

The horizontal component of `dvecE` is dE times the cosine of the angle between the `dvecE` and the horizontal, which equals, from the right triangle on the image, `x/sqrt(x^2 + a^2)` . Due to symmetry, all the vertical components of `dvecE` due to the differential charges around the ring will cancel, and only the horizontal components will add up. So,

`E = int dE = int k (dq)/(x^2 + r^2) * x/sqrt(x^2 + r^2) = kq_t x/(x^2+ r^2)^(3/2)`

The potential integral is then

`V(x) = -int_oo ^x kq_t x/(x^2 + r^2)^(3/2) dx`

This integral can be evaluated using the substitution `u = xdx` , and it results in

`V(x) = kq_t/sqrt(x^2 + r^2)` , same as the result found by superposition.