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A ring has charge Q and radius R. If a charge q is
placed at its centre, then increment in tension of ring

priya govind , 9 Years ago
Grade 12th pass
anser 1 Answers
Eshan

Last Activity: 7 Years ago

To understand how the tension in a charged ring changes when a charge is placed at its center, we need to delve into some fundamental principles of electrostatics and how they interact with forces in a physical system. Let's break this down step by step.

Understanding the Forces at Play

Consider a ring with a uniform charge \( Q \) distributed around its circumference. When a point charge \( q \) is placed at the center of this ring, it creates an electric field that influences the distribution of forces acting on the charged ring.

The Electric Field Inside the Ring

The electric field created by the charged ring at its center can be calculated using the principle of superposition. Each infinitesimal charge element \( dq \) on the ring contributes to the electric field at the center, but because of the symmetry, the horizontal components of these fields cancel out, leaving only a vertical component. However, at the very center, the net electric field due to the ring itself is zero.

Effect of the Central Charge

When we introduce a charge \( q \) at the center, this charge creates an electric field that exerts a force on the charged ring. The important part to note here is that the electric field \( E \) produced by the charge \( q \) at a distance \( R \) from itself (which is the radius of the ring) is given by:

\( E = \frac{k \cdot q}{R^2} \)

where \( k \) is Coulomb's constant. This electric field exerts a force \( F \) on each charge element of the ring:

\( F = dq \cdot E \)

Calculating the Increment in Tension

To find the increment in tension of the ring due to the central charge, we need to consider how this force translates into tension. The total force acting on the ring due to the electric field can be found by integrating over the entire charge of the ring:

  • First, calculate the force on one infinitesimal charge \( dq \) due to the electric field from \( q \).
  • This force will act radially outward and will affect the tension in the ring.
  • The total tension increase in the ring results from the vector sum of all these forces acting around the circumference.

Since the forces are symmetrically distributed, the total radial force \( F_{total} \) acting outward on the ring can be computed. The increment in tension \( \Delta T \) in the ring will then be directly proportional to this total radial force:

\( \Delta T = \frac{F_{total}}{2\pi R} \)

Final Expression for Increment in Tension

Putting everything together, if \( N \) is the total number of charge elements on the ring, the tension increment in the ring can be expressed as:

\( \Delta T = \frac{k \cdot q \cdot Q}{R^2} \)

This expression shows how the tension in the ring increases due to the presence of the charge \( q \) at its center, resulting from the Coulombic interactions between the charges.

In summary, adding a charge at the center of a charged ring leads to an increase in tension due to the electric force exerted by the central charge on the ring's charge distribution. This understanding is crucial in various applications, such as in electrostatic systems and designing charged particle accelerators.

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