To determine the tension created in a thin wire ring due to a point charge placed at its center, we need to analyze the forces acting on the ring. The ring carries a uniformly distributed charge, and when a point charge is placed at the center, it will exert an electric force on the charges in the ring. Let's break down the problem step by step.
Understanding the Setup
We have a thin ring with a radius \( R = 2.5 \) cm (which is \( 0.025 \) m) and a total charge \( Q = 5 \) microcoulombs (or \( 5 \times 10^{-6} \) C). A point charge \( Q_0 = 3 \) microcoulombs (or \( 3 \times 10^{-6} \) C) is placed at the center of the ring. The goal is to find the tension in the ring due to the electric field created by the point charge.
Electric Field Due to the Point Charge
The electric field \( E \) created by a point charge at a distance \( r \) from the charge is given by the formula:
- E = k \cdot \frac{|Q_0|}{r^2}
where \( k \) is Coulomb's constant, approximately \( 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \). Since the point charge is at the center of the ring, the distance \( r \) from the point charge to any point on the ring is equal to the radius \( R \).
Calculating the Electric Field
Substituting the values into the formula, we get:
- E = k \cdot \frac{|Q_0|}{R^2} = 8.99 \times 10^9 \cdot \frac{3 \times 10^{-6}}{(0.025)^2}
Calculating \( (0.025)^2 \) gives \( 0.000625 \) m². Now, substituting this value:
- E = 8.99 \times 10^9 \cdot \frac{3 \times 10^{-6}}{0.000625} = 8.99 \times 10^9 \cdot 4.8 \times 10^3
This results in:
- E \approx 4.32 \times 10^{13} \, \text{N/C}
Force on the Charges in the Ring
Each infinitesimal charge \( dq \) on the ring experiences a force due to the electric field \( E \). The force \( dF \) on a small segment of the ring can be expressed as:
Since the charge is uniformly distributed, we can express \( dq \) as:
- dq = \frac{Q}{2\pi R} \cdot d\theta
where \( d\theta \) is the infinitesimal angle subtended by the segment at the center of the ring.
Net Force and Tension in the Ring
The total force acting on the ring due to the electric field can be found by integrating \( dF \) around the entire ring. However, due to symmetry, the horizontal components of the forces will cancel out, and only the vertical components will contribute to the tension. The tension \( T \) in the ring can be thought of as balancing the net vertical force due to the electric field acting on the charge distribution.
Thus, the tension in the ring can be approximated as:
- T = \frac{Q \cdot E}{2} = \frac{5 \times 10^{-6} \cdot 4.32 \times 10^{13}}{2}
Calculating this gives:
- T \approx 1.08 \times 10^8 \, \text{N}
In summary, the tension created in the ring due to the presence of the point charge at its center is approximately \( 1.08 \times 10^8 \) N. This illustrates how electric forces can create significant tension in charged systems, especially when point charges are involved.
