Solution:

To understand why the magnitude of the electric field is \(E_0/2\) at four points on the axis of a uniformly charged ring, let's analyze this with the help of the electric field graph versus distance.

Key Concepts:
1. Electric Field on the Axis of a Uniformly Charged Ring:
The electric field at a distance \(x\) from the center of the ring along its axis is given by:
\[
E = \frac{k Q x}{(x^2 + R^2)^{3/2}}
\]
where:
- \(k\) is Coulomb's constant,
- \(Q\) is the total charge on the ring,
- \(R\) is the radius of the ring,
- \(x\) is the distance from the center along the axis.

2. Behavior of \(E\) as a Function of \(x\):
- At \(x = 0\): The electric field is \(0\) due to symmetry (no net field at the center).
- As \(x\) increases: \(E\) first increases, reaches a **maximum value** (\(E_0\)) at some distance \(x = x_{\text{max}}\), and then decreases asymptotically to \(0\) as \(x \to \infty\).

3. Points Where \(E = E_0/2\):
The equation \(E = \frac{E_0}{2}\) will have **two solutions** on either side of the point where \(E\) is maximum (\(x = \pm x_{\text{max}}\)):
- Two points closer to the ring's center (on both sides of \(x = 0\)).
- Two points farther from the center (on both sides of \(x = 0\)).

Thus, there are a total of four points where \(E = \frac{E_0}{2}\).

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Graphical Explanation:
1. The graph of \(E\) versus \(x\) is symmetric about \(x = 0\) and resembles a bell-shaped curve.
2. \(E\) starts at \(0\) at \(x = 0\), increases to a peak value of \(E_0\) at \(x_{\text{max}}\), and then decreases symmetrically as \(x\) moves away from \(x_{\text{max}}\) on both sides.
3. To find the points where \(E = \frac{E_0}{2}\), draw a horizontal line at \(E = \frac{E_0}{2}\). This line will intersect the \(E\)-vs-\(x\) curve at **four points**:
- Two on the rising part of the curve (closer to \(x = 0\)),
- Two on the falling part of the curve (farther from \(x = 0\)).

Conclusion:
The electric field is \(E_0/2\) at four points on the axis of the ring—two on each side of the ring's center. These correspond to the solutions of the equation \(E = \frac{k Q x}{(x^2 + R^2)^{3/2}} = \frac{E_0}{2}\).