A ring of radius R has charge `-Q` distributed uniformly over it. Calculate the charge that should be placed at the center of the ring such that the electric field becomes zero at apoint on the axis of the ring at distant R from the center of the ring.

Explanation

To solve the problem, we need to find the charge that should be placed at the center of a ring with a uniform negative charge distribution, such that the electric field at a point on the axis of the ring becomes zero.

### Step-by-Step Solution:

1. **Understand the Configuration**:
- We have a ring of radius \( R \) with a uniform negative charge \( -Q \) distributed over it.
- We need to find a positive charge \( q \) to be placed at the center of the ring such that the electric field at a point \( P \) on the axis of the ring, at a distance \( R \) from the center, is zero.

2. **Electric Field due to the Ring**:
- The electric field \( E_{\text{ring}} \) at a point on the axis of the ring (distance \( x \) from the center) is given by the formula:
\[ E_{\text{ring}} = \frac{kQx}{(R^2 + x^2)^{3/2}} \]
- Here, \( k \) is Coulomb's constant, \( Q \) is the total charge on the ring, and \( x \) is the distance from the center of the ring to point \( P \).

3. **Substituting Values**:
- In our case, \( x = R \) (the distance from the center to point \( P \)).
- Therefore, substituting \( x = R \) into the formula gives:
\[ E_{\text{ring}} = \frac{k(-Q)R}{(R^2 + R^2)^{3/2}} = \frac{k(-Q)R}{(2R^2)^{3/2}} = \frac{k(-Q)R}{(2\sqrt{2}R^3)} = \frac{-kQ}{2\sqrt{2}R^2} \]

4. **Electric Field due to the Charge at Center**:
- The electric field \( E_q \) due to the charge \( q \) at the center of the ring at point \( P \) (distance \( R \) away) is given by:
\[ E_q = \frac{kq}{R^2} \]
- This electric field will be directed away from the positive charge \( q \).

5. **Setting the Electric Fields Equal**:
- For the total electric field at point \( P \) to be zero, the electric field due to the ring must equal the electric field due to the charge at the center:
\[ E_{\text{ring}} + E_q = 0 \]
- Substituting the expressions we derived:
\[ \frac{-kQ}{2\sqrt{2}R^2} + \frac{kq}{R^2} = 0 \]

6. **Solving for Charge \( q \)**:
- Rearranging the equation gives:
\[ \frac{kq}{R^2} = \frac{kQ}{2\sqrt{2}R^2} \]
- Canceling \( k \) and \( R^2 \) (assuming \( R \neq 0 \)):
\[ q = \frac{Q}{2\sqrt{2}} \]

7. **Final Answer**:
- The charge that should be placed at the center of the ring is:
\[ q = \frac{Q}{2\sqrt{2}} \]