A circular wire loop of radius R carries a total charge q distributed uniformly over its length. A small length `x (ltlt R)` of the wire is cut off. Find the electric field at the centre due to the remaining wire.

Explanation

To find the electric field at the center of a circular wire loop after cutting off a small length \( x \) from it, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have a circular wire loop of radius \( R \) carrying a total charge \( q \) uniformly distributed along its length.
- A small length \( x \) (where \( x \ll R \)) is cut off from the loop.
- We need to find the electric field at the center of the remaining wire.

2. **Electric Field of the Complete Loop**:
- The electric field at the center of a uniformly charged circular loop is zero. This is because the contributions to the electric field from all the infinitesimal charge elements cancel out due to symmetry.

3. **Calculating the Charge on the Cut Length**:
- The total length of the wire loop is \( 2\pi R \).
- The charge per unit length \( \lambda \) on the wire is given by:
\[ \lambda = \frac{q}{2\pi R} \]
- The charge \( q' \) on the cut length \( x \) can be calculated as:
\[ q' = \lambda \cdot x = \frac{q}{2\pi R} \cdot x \]

4. **Electric Field Due to the Cut Length**:
- The electric field \( E_2 \) at the center due to the cut length can be calculated using Coulomb's law:
\[ E_2 = k \frac{q'}{R^2} \]
- Substituting \( q' \):
\[ E_2 = k \frac{\frac{q}{2\pi R} \cdot x}{R^2} = \frac{kqx}{2\pi R^3} \]
- Here, \( k \) is Coulomb's constant.

5. **Electric Field Due to the Remaining Wire**:
- Since the electric field due to the entire loop is zero, the electric field \( E_1 \) due to the remaining part of the wire must be equal in magnitude but opposite in direction to \( E_2 \):
\[ E_1 = -E_2 \]
- Therefore, the electric field at the center due to the remaining wire is:
\[ E_1 = -\frac{kqx}{2\pi R^3} \]

### Final Result:
The electric field at the center of the remaining wire loop after cutting off a small length \( x \) is:
\[ E = -\frac{kqx}{2\pi R^3} \]