To find the y-component of the electric field at point P due to a uniformly charged circular ring, let's break it down step by step.

Problem Overview:

• A uniformly charged ring with total charge $Q$ and radius $R$ lies in the $y$-$z$-plane (i.e., at $x = 0$).
• Point $P$ is located on the $x$-axis at a distance $d$ from the origin, i.e., at coordinates $(d, 0, 0)$.
• Coulomb’s constant is $k_e$.

Step 1: Electric Field due to a Small Element of Charge

First, let's consider a small charge element $dq$ on the ring. The total charge is uniformly distributed along the ring. Hence, the linear charge density is given by:

$\lambda = \frac{Q}{2\pi R}$

Let’s take a small element of charge $dq$ at an angle $\theta$ in the $y$-$z$-plane. The position of this element is given by:

$(x_{\text{elem}}, y_{\text{elem}}, z_{\text{elem}}) = (R \cos \theta, R \sin \theta, 0)$

This element of charge exerts a Coulomb force at point $P$. The distance from the charge element to point $P$ (which is at $(d, 0, 0)$) is:

$r = \sqrt{(d - R \cos \theta)^2 + (R \sin \theta)^2}$

Simplifying this:

$r = \sqrt{(d - R \cos \theta)^2 + R^2 \sin^2 \theta} = \sqrt{d^2 - 2dR \cos \theta + R^2}$

The electric field due to this small charge element $dq$ is given by Coulomb’s law:

$d\mathbf{E} = \frac{k_e \, dq}{r^2} \hat{r}$

where $\hat{r}$ is the unit vector pointing from the charge element to point $P$.

Step 2: Symmetry Considerations

Due to the symmetry of the problem, we can observe that the electric field components in the $y$- and $z$-directions will cancel out. This is because for every element at an angle $\theta$, there is a corresponding element at an angle $\theta + \pi$, resulting in cancellation of the vertical components (the $y$ and $z$ components). Therefore, only the $x$-component of the electric field will remain.

Step 3: Contribution to the Electric Field in the $y$-Direction

Since we are interested in the $y$-component of the electric field at point $P$, and from the symmetry argument, we conclude that the $y$-component of the electric field is zero.

This is because for every charge element creating a field in the positive $y$-direction, there is a corresponding element creating a field in the negative $y$-direction, and the contributions will cancel out exactly.

Final Answer:

The $y$-component of the electric field at point $P$ is:

$E_y = 0$

Would you like further clarification on any part of the problem or its solution? Here are some related questions for deeper exploration:

1. What is the electric field at point $P$ in the $x$-direction?
2. How does the electric field change if point $P$ is placed off the $x$-axis?
3. How would the electric field at point $P$ change if the charge on the ring were distributed non-uniformly?
4. What happens to the electric field at point $P$ as $d \to \infty$?
5. How would you derive the electric field for a ring with a negative total charge?

Tip: When working with symmetry in electrostatic problems, always consider how the charge distribution affects different components of the field. Symmetry often simplifies calculations by eliminating components of the field that must cancel out.