Question

Six identical charged particles having charge q are fixed at the vertices of a regular hexagon of side a. A negative charge –Q having mass m is situated at the centre of the hexagon. The minimum velocity given to charge –Q so that it escape from the charge system, is

Answer 1

Given:

Six identical charged particles having charge q are fixed at the vertices of a regular hexagon of side a. A negative charge –Q having mass m is situated at the centre of the hexagon.

To find:

Minimum escape velocity for -Q in order to escape from the charge system.

Calculation:

We have to supply a kinetic energy (to -Q charge) equivalent to the total potential energy of the charge system. This would allow the -Q charge to leave the system.

$\sf{ \therefore \: KE = PE}$

$\sf{ = > \: \dfrac{1}{2} m {v}^{2} = 6 \times ( \dfrac{kQq}{a} )}$

$\sf{ = > \: m {v}^{2} = 12 \times ( \dfrac{kQq}{a} )}$

$\sf{ = > \: {v}^{2} = 12 \times ( \dfrac{kQq}{ma} )}$

$\sf{ = > \: v = \sqrt{12 \times ( \dfrac{kQq}{ma} )}}$

Putting value of Coulomb's Constant:

$\sf{ = > \: v = \sqrt{12 \times ( \dfrac{Qq}{4\pi \epsilon_{0} ma} )}}$

$\sf{ = > \: v = \sqrt{ \dfrac{3Qq}{\pi \epsilon_{0} ma} }}$

So, final answer is:

$\boxed{ \large{ \red{ \rm{ \: v = \sqrt{ \dfrac{3Qq}{\pi \epsilon_{0} ma} }}}}}$