The maximum electric field at a point on the axis of a uniformly charged ring is `E_(0)`. At how many points on the axis will the magnitude of the electric field be `E_(0)//2`.

Options

A. 1
B. 2
C. 3
D. 4

Explanation

To solve the problem of finding how many points on the axis of a uniformly charged ring have an electric field of magnitude \( \frac{E_0}{2} \), we can follow these steps:

### Step 1: Understand the Electric Field of a Charged Ring
The electric field \( E \) at a point on the axis of a uniformly charged ring is given by the formula:
\[ E = \frac{1}{4 \pi \epsilon_0} \frac{Q x}{(R^2 + x^2)^{3/2}} \]
where:
- \( Q \) is the total charge on the ring,
- \( R \) is the radius of the ring,
- \( x \) is the distance from the center of the ring along the axis.

### Step 2: Determine the Maximum Electric Field
The maximum electric field \( E_0 \) occurs at a specific point on the axis, which we can find by differentiating the electric field expression with respect to \( x \) and setting the derivative to zero. The condition for maximum electric field gives:
\[ x = \frac{R}{\sqrt{2}} \]
At this point, we can calculate the maximum electric field \( E_0 \).

### Step 3: Set Up the Equation for \( \frac{E_0}{2} \)
We need to find the points where the electric field equals \( \frac{E_0}{2} \):
\[ \frac{1}{4 \pi \epsilon_0} \frac{Q x}{(R^2 + x^2)^{3/2}} = \frac{E_0}{2} \]
Substituting \( E_0 \) into the equation, we can simplify it.

### Step 4: Solve for \( x \)
Rearranging the equation gives:
\[ \frac{Q x}{(R^2 + x^2)^{3/2}} = 2 \cdot \frac{E_0}{4 \pi \epsilon_0} \]
Let’s denote \( k = \frac{1}{4 \pi \epsilon_0} \), then:
\[ \frac{Q x}{(R^2 + x^2)^{3/2}} = 2 k \cdot \frac{Q}{R^2} \]
This simplifies to:
\[ \frac{x}{(R^2 + x^2)^{3/2}} = \frac{2}{R^2} \]

### Step 5: Find the Roots of the Equation
To find the points where the electric field is \( \frac{E_0}{2} \), we can solve the equation:
\[ x^2 = \frac{2(R^2 + x^2)^{3/2}}{R^2} \]
This will yield a polynomial equation in \( x \). Solving this will give us the values of \( x \).

### Step 6: Analyze the Solutions
After solving the polynomial, we will find that there are four points on the axis where the electric field has the magnitude \( \frac{E_0}{2} \). Two points will be located symmetrically on either side of the maximum point and two points will be further away.

### Conclusion
Thus, the number of points on the axis where the electric field is \( \frac{E_0}{2} \) is **4**.

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