A ring of radius 0.1m is made out of thin metallic wire of area of cross section `10^(-6)m^(2)`. The ring has a uniform charge of `pi` coulombs. Find the change in the radius of the rig when a charge of `10^(-8)C` is placed at the center of the ring. Young's modulus of hte metal is `2xx10^(11)Nm^(-2)`.

Explanation

To solve the problem step by step, we will follow the outlined approach based on the information given in the question.

### Step 1: Understand the problem
We have a ring of radius \( r = 0.1 \, \text{m} \) with a uniform charge \( Q = \pi \, \text{C} \). A charge \( q = 10^{-8} \, \text{C} \) is placed at the center of the ring. We need to find the change in the radius of the ring when this charge is introduced, given the Young's modulus \( Y = 2 \times 10^{11} \, \text{N/m}^2 \).

### Step 2: Calculate the charge per unit length (λ)
The total length \( L \) of the ring can be calculated using the formula for the circumference of a circle:
\[ L = 2\pi r = 2\pi(0.1) = 0.2\pi \, \text{m} \]
Now, we can find the charge per unit length \( \lambda \):
\[ \lambda = \frac{Q}{L} = \frac{\pi}{0.2\pi} = \frac{1}{0.2} = 5 \, \text{C/m} \]

### Step 3: Determine the force on a small arc of the ring
Consider a small arc of the ring subtending an angle \( \theta \) at the center. The force \( F \) on this arc due to the charge \( q \) at the center can be calculated using Coulomb's law. The force on the arc can be expressed as:
\[ F = k \cdot \frac{q \cdot \lambda \cdot \theta \cdot r}{r^2} \]
where \( k \) is Coulomb's constant \( (k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2) \).

### Step 4: Calculate the tension in the wire
The tension \( T \) in the wire can be related to the force acting on the small arc. The vertical components of the tension must balance the force due to the charge at the center:
\[ F = 2T \sin\left(\frac{\theta}{2}\right) \approx T \theta \quad (\text{for small } \theta) \]
Thus, we have:
\[ T = \frac{F}{\theta} \]

### Step 5: Substitute the values into the tension formula
Substituting the expression for \( F \):
\[ T = \frac{k \cdot q \cdot \lambda \cdot r}{r^2 \cdot \theta} \]
Now, substituting \( \lambda = 5 \, \text{C/m} \), \( q = 10^{-8} \, \text{C} \), and \( r = 0.1 \, \text{m} \):
\[ T = \frac{9 \times 10^9 \cdot 10^{-8} \cdot 5 \cdot 0.1}{(0.1)^2 \cdot \theta} \]
This simplifies to:
\[ T = \frac{4500}{\theta} \, \text{N} \]

### Step 6: Calculate the change in radius using Young's modulus
Using the definition of Young's modulus:
\[ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{T/A}{\Delta R/R} \]
Where \( A = 10^{-6} \, \text{m}^2 \) is the cross-sectional area of the wire. Rearranging gives:
\[ \Delta R = \frac{T \cdot R}{Y \cdot A} \]
Substituting \( T = 4500 \, \text{N} \), \( R = 0.1 \, \text{m} \), \( Y = 2 \times 10^{11} \, \text{N/m}^2 \), and \( A = 10^{-6} \, \text{m}^2 \):
\[ \Delta R = \frac{4500 \cdot 0.1}{2 \times 10^{11} \cdot 10^{-6}} = \frac{450}{2 \times 10^5} = 2.25 \times 10^{-3} \, \text{m} = 2.25 \, \text{mm} \]

### Final Answer
The change in the radius of the ring is \( \Delta R = 2.25 \, \text{mm} \).

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