A ring with charge \(Q\) and radius \(R\) is in the \(y z\) -plane and centered on the origin. What is the electric potential a distance \(x\) above the center of the ring? Derive the electric field from this relationship.

Let's consider a point P located along the x-axis at a distance x above the center of the ring. The electric potential at point P is given by the following equation: \(V(P) = \dfrac{1}{4\pi\epsilon_0} \int \dfrac{dq}{r}\) where \(1/4\pi\epsilon_0\) is the electrostatic constant, dq represents an infinitesimal charge element on the ring, and r is the distance from the charge element to point P. To find the electric potential, we will need to integrate over all the dq charges on the ring, considering the geometry of the problem.

Let's define an infinitesimal angle \(\delta \phi\) in the yz-plane, which would correspond to the infinitesimal arc length \(\delta l = R \delta \phi\), and hence, the infinitesimal charge element dq would be \(\delta q = \dfrac{Q}{2\pi R} \delta l = \dfrac{Q}{2\pi R} R \delta \phi = \dfrac{Q}{2\pi} \delta \phi\). The distance r from the charge element \(\delta q\) to the point P can be determined by applying the Pythagorean theorem: \(r = \sqrt{x^2 + R^2}\) Now we can set up the integral for the electric potential at point P: \(V(P) = \dfrac{1}{4\pi\epsilon_0} \int \dfrac{dq}{r} = \dfrac{1}{4\pi\epsilon_0} \int \dfrac{\dfrac{Q}{2\pi}\delta\phi}{\sqrt{x^2+R^2}}\)

The integral we obtained is: \(V(P) = \dfrac{1}{4\pi\epsilon_0} \int \dfrac{Q}{2\pi(x^2+R^2)^{1/2}}\delta\phi\) Since x and R are constants, we can move them outside the integral, and integrate with respect to \(\phi\): \(V(P) = \dfrac{Q}{4\pi\epsilon_0} \dfrac{1}{(x^2+R^2)^{1/2}} \int_0^{2\pi}\delta\phi\) After evaluating the integral, we get: \(V(P) = \dfrac{Q}{4\pi\epsilon_0} \dfrac{1}{\sqrt{x^2+R^2}} \cdot 2\pi = \dfrac{Q}{2\epsilon_0} \dfrac{1}{\sqrt{x^2+R^2}}\)

The electric field can be found as the negative gradient of the electric potential. In this case, the only non-zero component of the electric field is along the x-axis, so we'll calculate the derivative of the electric potential with respect to x: \(E_x = -\dfrac{dV}{dx} = -\dfrac{d}{dx} \left(\dfrac{Q}{2\epsilon_0} \dfrac{1}{\sqrt{x^2+R^2}} \right)\) After taking the derivative and simplifying, we get: \(E_x = \dfrac{Q}{2\epsilon_0} \dfrac{x}{(x^2+R^2)^{3/2}}\) Therefore, the electric field at point P has only an x-component and can be written as: \(\textbf{E} = \left(\dfrac{Q}{2\epsilon_0} \dfrac{x}{(x^2+R^2)^{3/2}}\right) \hat{\textbf{x}}\) The electric potential at a distance x above the center of the charged ring is \(V(P) = \dfrac{Q}{2\epsilon_0} \dfrac{1}{\sqrt{x^2+R^2}}\), and the electric field derived from this relationship is \(\textbf{E} = \left(\dfrac{Q}{2\epsilon_0} \dfrac{x}{(x^2+R^2)^{3/2}}\right) \hat{\textbf{x}}\).