A thin circular wire of radius r has a charge Q. If a point charge q is placed at the centre of the ring, then find the increase in tension in the wire.

Explanation

To find the increase in tension in a thin circular wire of radius \( r \) with a charge \( Q \) when a point charge \( q \) is placed at the center of the ring, we can follow these steps:

### Step 1: Define Linear Charge Density
The linear charge density \( \lambda \) can be defined as the total charge \( Q \) divided by the circumference of the ring.
\[ \lambda = \frac{Q}{2\pi r} \]

### Step 2: Consider a Small Section of the Wire
Consider a small section \( dx \) of the wire. The tension in the wire will be affected by the repulsive force due to the point charge \( q \) at the center.

### Step 3: Calculate the Force on the Small Section
Using Coulomb's law, the force \( F \) acting on the small section \( dx \) due to the point charge \( q \) can be expressed as:
\[ F = k \cdot q \cdot \lambda \cdot \frac{dx}{r^2} \]
where \( k \) is Coulomb's constant.

### Step 4: Resolve Tension Components
The tension \( T \) in the wire acts tangentially to the circular wire. If we denote the angle subtended by the small section \( dx \) as \( \theta \), we can resolve the tension into components. The vertical components of tension from two adjacent sections will balance the force \( F \):
\[ F = 2T \sin\left(\frac{\theta}{2}\right) \]

### Step 5: Approximate for Small Angles
For small angles, we can use the approximation \( \sin\left(\frac{\theta}{2}\right) \approx \frac{\theta}{2} \). Thus, we can rewrite the force balance as:
\[ F = T \cdot \theta \]

### Step 6: Relate Arc Length to Angle
The arc length \( dx \) can be expressed in terms of the angle \( \theta \):
\[ dx = r \cdot \theta \]

### Step 7: Substitute and Solve for Tension
Substituting \( dx \) into the expression for force:
\[ F = k \cdot q \cdot \lambda \cdot \frac{r \cdot \theta}{r^2} = \frac{k \cdot q \cdot \lambda \cdot \theta}{r} \]
Setting the two expressions for \( F \) equal gives:
\[ T \cdot \theta = \frac{k \cdot q \cdot \lambda \cdot \theta}{r} \]
Cancelling \( \theta \) (assuming \( \theta \neq 0 \)):
\[ T = \frac{k \cdot q \cdot \lambda}{r} \]

### Step 8: Substitute for \( \lambda \)
Substituting \( \lambda = \frac{Q}{2\pi r} \):
\[ T = \frac{k \cdot q \cdot \frac{Q}{2\pi r}}{r} = \frac{k \cdot q \cdot Q}{2\pi r^2} \]

### Step 9: Final Expression for Tension
Using \( k = \frac{1}{4\pi \epsilon_0} \):
\[ T = \frac{q \cdot Q}{8\pi^2 \epsilon_0 r^2} \]

Thus, the increase in tension in the wire due to the point charge \( q \) at the center is:
\[ \Delta T = \frac{q \cdot Q}{8\pi^2 \epsilon_0 r^2} \]