The diagram shows a small bead carrying charge q. The bead can freely move on smooth fixed ring placed on a smooth horizontal plane. Potential due to + Q at 1 is V. If bead is given a very small velocity tangential from Q Velocity of bead when it reaches at 1 (mass of bead is m )

Question

The diagram shows a small bead carrying charge q. The bead can freely move on smooth fixed ring placed on a smooth horizontal plane. Potential due to + Q at 1 is V. If bead is given a very small velocity tangential from Q Velocity of bead when it reaches at 1 (mass of bead is m ) <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/756e7bbeef3e2337475ef2c16d998d79-.png" /> (A) √(6 qV / m ) (B) √(q V/m) (C) √(3 qV / m ) (D) Zero

Tardigrade Admin · Accepted Answer

As between (a) and (b) Δ v =3 V ⇒ work =Δ KE ⇒ 3 qV =(1/2) mv 2 ⇒ v=√(6 q V/m) <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/d4a76f7665a378f8616053637f9d14e4-.png" />

Q. The diagram shows a small bead carrying charge $q .$ The bead can freely move on smooth fixed ring placed on a smooth horizontal plane. Potential due to $+ Q$ at 1 is $V .$ If bead is given a very small velocity tangential from $Q$ Velocity of bead when it reaches at $1$ (mass of bead is $m$ )

$\frac{6 q V}{m}$

$\frac{q V}{m}$

$\frac{3 q V}{m}$

Zero

As between $(a)$ and $(b)$
$Δ v = 3 V$
$\Rightarrow$ work $= Δ K E$
$\Rightarrow 3 q V = \frac{1}{2} m v^{2}$
$\Rightarrow v = \frac{6 q V}{m}$

Q. The diagram shows a small bead carrying charge q. The bead can freely move on smooth fixed ring placed on a smooth horizontal plane. Potential due to +Q at 1 is V. If bead is given a very small velocity tangential from Q Velocity of bead when it reaches at 1 (mass of bead is m )

m6qV​​

mqV​​

m3qV​​