# Tension developed in a charged ring

• Hamiltonian
In summary, consider a small element that subtends an angle ##2\Delta \theta## at the center of the ring. balancing the forces on this element gives:2T\Delta \theta = E(dq) = E (\frac{Q}{2\pi})(2\Delta \theta)
Hamiltonian
Homework Statement
calculate the tension developed in ring of radius ##R##( of negligible thickness) and charge ##Q##
Relevant Equations
-

consider a small element that subtends an angle ##2\Delta \theta## at the center of the ring. balancing the forces on this element gives:
(let the field due to the ring be at its circumference be ##E##).
$$2T\Delta \theta = E(dq) = E (\frac{Q}{2\pi})(2\Delta \theta)$$
$$T = \frac{EQ}{2\pi}$$
now the problem is reduced to finding the field due to the charged ring at its circumference:

let the distance from an infinitesimal charge ##(dq)## to the required point on the circumference be ##z##
$$z^2 = (Rsin\theta + R)^2 + (Rcos\theta)^2 = 2R^2(1+sin\theta)$$
$$(dE) = k(dq)/z^2 = \frac {k(dq)}{2R^2(1+sin\theta)}$$
by symmetry arguments the net field at the circumference will be only in the y direction:
$$(dE_{net}) = (dE)(cos\alpha)$$
$$\alpha = \pi/4 - \theta/2$$
$$(dq) = (Q/2\pi) d\theta$$
$$dE_{net} = k(dq)\frac{cos(\pi/4 - \theta/2)}{1+sin\theta} = \frac{kQ}{4\pi R^2} \frac{cos(\pi/4 - \theta/2)}{1+sin\theta} (d\theta)$$
$$E_{net} = \frac{kQ}{4\pi R^2} \int_0^{2\pi} \frac{cos(\pi/4 - \theta/2)}{1+sin\theta} d\theta$$

I put the integral into wolframalpha and it does not converge!
that would mean the field is infinite which obviously can't be true.
is there any physical significance to this or have i made a mistake in calculating the field at the circumference of the ring.

I had also tried to find the field at any general point in the plane of the ring and then wanted to find the field at circumference but the integral proved to be very complex (wolframalpha exceeded standard computation time) hence i didn't go down that path.

And a point charge !

## 1. What is tension developed in a charged ring?

Tension developed in a charged ring refers to the force that is exerted on the ring due to the presence of electric charges.

## 2. How is tension developed in a charged ring calculated?

Tension developed in a charged ring can be calculated using the formula T = (kQ2) / r, where T is the tension, k is the Coulomb's constant, Q is the charge on the ring, and r is the radius of the ring.

## 3. What factors affect the tension developed in a charged ring?

The tension developed in a charged ring is affected by the amount of charge on the ring, the distance between the charges, and the radius of the ring. It is also affected by the medium in which the ring is placed, as different mediums have different dielectric constants.

## 4. How does tension in a charged ring affect its stability?

The tension developed in a charged ring plays a crucial role in determining its stability. If the tension is too high, the ring may break due to the repulsive forces between the charges. On the other hand, if the tension is too low, the ring may lose its shape and collapse.

## 5. Can tension in a charged ring be used for any practical applications?

Yes, tension developed in a charged ring has practical applications in various fields such as electrostatic generators, particle accelerators, and ion traps. It is also used in research and experiments related to electromagnetism and plasma physics.

• Introductory Physics Homework Help
Replies
10
Views
1K
• Introductory Physics Homework Help
Replies
12
Views
426
• Introductory Physics Homework Help
Replies
99
Views
9K
• Introductory Physics Homework Help
Replies
5
Views
507
• Introductory Physics Homework Help
Replies
2
Views
1K
• Introductory Physics Homework Help
Replies
2
Views
786
• Introductory Physics Homework Help
Replies
6
Views
2K
• Introductory Physics Homework Help
Replies
3
Views
1K
• Introductory Physics Homework Help
Replies
9
Views
1K
• Introductory Physics Homework Help
Replies
9
Views
2K