Question

A particle having charge -q and mass m is
released from rest on the axis of a fixed ring of
total charge Q and radius R from a distance V3R.
Its kinetic energy when it reaches the centre of
ring is​

Answer 1

Answer: kQq/2R         ($\sqrt{R^{2} + \sqrt{3^{2} }R^{2} }$)

Explanation :lets take position of particle A then P.E. at A is -kQq/2R and K.E. = 0

now at the centre of the ring P.E. id -kQq/R and Let  K.E. at centre of ring  = x

with evergy conservation

-kQq/2R + 0 = -kQq/R + x

x = kQq/2R

Answer 2

Answer: Kinetic energy of -q charge when it reaches the centre of ring is​  kQq/2R

Explanation:

Initial condition ⇒ -q charge at √3R distance from centre

Initial Kinetic Energy = 0

Initial Potential Energy = $\frac{-kQq}{\sqrt{R^{2} + (\sqrt{3}R) ^{2} } }$  =  $\frac{-kQq}{2R}$

Total initial energy = $\frac{-kQq}{2R}$

Final condition ⇒ -q charge at centre

Let final Kinetic Energy be $x$

Final Potential Energy = $\frac{-kQq}{R}$

Total final energy = $\frac{-kQq}{R}$ $+ x$

Applying energy conservation,

Initial energy =  Final energy

$\frac{-kQq}{2R}$   =  $\frac{-kQq}{R}$ $+ x$

$x = \frac{kQq}{2R}$

Kinetic energy at centre is $\frac{kQq}{2R}$

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