A ring is positively charge with uniform linear charge density \(\lambda\) and radius \(r\) , now a positive charge \(q_0\) is placed at the center of the ring in the same plane.

Find the total tension after \( q_0\) is placed.

Permitivity of free space is \(\epsilon _0 \).

Hint: There will be tension due to two.

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestyes , tanishq's answer is absolutely correct ! ( i guess so )

Log in to reply

I guess this is the answer.. T=kQq / 2.pi.R^2 But can you solve it and show..

Log in to reply

Is it correct

\(T=\frac{\lambda q_{0}}{4\pi \epsilon_{o} r}\)

Log in to reply

It is not as simple you think .. If so then I'am sure Kushal don't post it .. Since that was very easy and standard problem .. :D

Log in to reply

Total tension will be developed due to electrostatic force by \(q_0\) and the

ring.If the question was to find the increase in tension when \(q_o\) was placed then your answer is correct.

Log in to reply