Question

Expansion of Ring due to Charge: A ring of radius 0.1 m is made out of a thin metallic wire of area of cross section $10^{-6} m^2$. The ring has a uniform charge of $\pi$ coulomb. Find the change in radius of ring when a charge of $10^{-8}$ C is placed at the centre of ring. Given that Young's modulus of the metal is $2 \times 10^{11} N/m^2$.

Answer 1

$\Delta R = 2.25 \times 10^{-3}$ m (Expansion in radius)

Answer 2

Solution:

1. Electric Field by the Central Charge:
   The electric field at the ring (radius $R=0.1$ m) due to the central charge $Q_c=10^{-8}$ C is

   $$E = \frac{k Q_c}{R^2} = \frac{9\times10^9 \times 10^{-8}}{(0.1)^2} = \frac{90}{0.01} = 9000\ \text{N/C}.$$

2. Force per Unit Length on the Ring:
   The ring carries a uniform total charge $Q_{\text{ring}} = \pi$ C. Therefore, the linear charge density is

   $$\lambda = \frac{Q_{\text{ring}}}{2\pi R} = \frac{\pi}{2\pi \times 0.1} = \frac{1}{0.2} = 5\ \text{C/m}.$$

   Thus, the force per unit length on the ring is

   $$f = \lambda E = 5 \times 9000 = 45000\ \text{N/m}.$$

3. Tension in the Ring:
   Consider a small segment subtending angle $d\theta$. Tension $T$ in the ring provides an inward force:

   $$\text{Inward force} \approx T\,d\theta.$$

   This must balance the outward force on the segment (with arc length $R\,d\theta$):

   $$T\,d\theta = f\,R\,d\theta \quad\Longrightarrow\quad T = fR = 45000 \times 0.1 = 4500\ \text{N}.$$

4. Expansion of the Ring:
   The total length of the ring is $L = 2\pi R$. Under tension, the extension is given by

   $$\Delta L = \frac{T L}{AY},$$

   where $A = 10^{-6}\ \text{m}^2$ is the cross-sectional area and $Y = 2\times10^{11}\ \text{N/m}^2$ is Young’s modulus. The corresponding change in radius is:

   $$\Delta R = \frac{\Delta L}{2\pi} = \frac{T R}{AY}.$$

   Substituting the values:

   $$\Delta R = \frac{4500 \times 0.1}{10^{-6} \times 2\times10^{11}} = \frac{450}{2\times10^{5}} = \frac{450}{200000} = 0.00225\ \text{m}.$$

Explanation (Minimal):

• Compute $E = \frac{kQ_c}{R^2}$.
• Find linear charge density $\lambda = \frac{\pi}{2\pi(0.1)} = 5\ \text{C/m}$.
• Force per unit length $f = \lambda E = 45000\ \text{N/m}$.
• Tension $T = fR = 4500\ \text{N}$.
• Extension: $\Delta R = \frac{T R}{AY} = 0.00225$ m.