Tension in a charged ring.

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.

#ElectricityAndMagnetism

Easy Math Editor

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
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`\sum_{i=1}^3`	$\sum_{i=1}^3$
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Comments

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

A Former Brilliant Member - 4 years, 6 months ago

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

Dibyanshu Patnaik - 5 years, 7 months ago

Is it correct

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

Tanishq Varshney - 6 years, 1 month ago

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

Karan Shekhawat - 6 years, 1 month ago

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.

Kushal Patankar - 6 years, 1 month ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$