a circular ring carries a charge` Q_1` the variation of electric field with distance x measured from centre along Axis for x>>R can be given as (R radius of ring)

Options

A. `Eprop1/x^3`
B. `Epropx`
C. `Eprop1/x^2`
D. `Eprop1/x`

Answer

To solve the problem of how the electric field \( E \) varies with distance \( x \) from the center of a circular ring carrying a charge \( Q_1 \) along its axis for \( x \gg R \) (where \( R \) is the radius of the ring), we can follow these steps:

### Step 1: Understand the Configuration
We have a circular ring of radius \( R \) with a total charge \( Q_1 \) uniformly distributed along its circumference. We want to find the electric field at a point \( P \) located on the axis of the ring at a distance \( x \) from its center.

### Step 2: Write the Electric Field Expression
The electric field \( E \) at a point along the axis of the ring can be expressed using the formula:

\[ E = \frac{k Q_1 x}{(R^2 + x^2)^{3/2}} \]

where \( k \) is Coulomb's constant, \( Q_1 \) is the charge on the ring, \( R \) is the radius of the ring, and \( x \) is the distance from the center of the ring to the point where the electric field is being calculated.

### Step 3: Analyze the Condition \( x \gg R \)
Since we are interested in the case where \( x \) is much greater than \( R \) (i.e., \( x \gg R \)), we can simplify the expression. In this case, \( R^2 \) becomes negligible compared to \( x^2 \):

\[ R^2 + x^2 \approx x^2 \]

### Step 4: Simplify the Electric Field Expression
Substituting this approximation into the electric field expression gives:

\[ E \approx \frac{k Q_1 x}{(x^2)^{3/2}} = \frac{k Q_1 x}{x^3} \]

### Step 5: Further Simplification
Now, simplifying this expression:

\[ E \approx \frac{k Q_1}{x^2} \]

### Step 6: Conclusion
From the final expression, we see that the electric field \( E \) is inversely proportional to the square of the distance \( x \):

\[ E \propto \frac{1}{x^2} \]

Thus, the variation of the electric field with distance \( x \) for \( x \gg R \) is given by:

\[ E \propto \frac{1}{x^2} \]

### Final Answer
The electric field \( E \) varies as \( \frac{1}{x^2} \) when \( x \) is much greater than the radius \( R \) of the ring.

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