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Figure shows two large cylindrical shells having uniform linear charge densities $+\lambda$ and $-\lambda$ . Radius of inner cylinder is ‘ $a$ ’ and that of outer cylinder is ‘ $b$ ’. A charged particle of mass $m$ , charge $q$ revolves in a circle of radius $r$ (where $a$ < $r$ < $b$ ). Then it's speed ‘ $V$ ’ is : (Neglect gravity and assume the radii of both the cylinders to be very small in comparison

to their length.
Option: 1
$\sqrt{\frac{\lambda q}{2\pi \in _{0}m}}$

Option: 2
$\sqrt{\frac{2\lambda q}{\pi \in _{0}m}}$

Option: 3
$\sqrt{\frac{\lambda q}{\pi \in _{0}m}}$

Option: 4
$\sqrt{\frac{\lambda q}{4\pi \in _{0}m}}$

Answers (1)

best_answer

The electric field between the two cylinders $=\frac{2k\lambda }{r}$

Force on charge $q=\frac{2k\lambda q}{r}$

This force is equal to a centripetal force

$\frac{2k\lambda q}{r}= \frac{mv^{2}}{r}$

So $V=\sqrt{2\frac{1\lambda q}{4\pi \in _{0}m}}=\sqrt{\frac{\lambda q}{2\pi \in _{0}m}}$

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