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

ANSWER

force between Q and q=F[F=KQq/r

2

]

force between Q and -2q=−2KQq/r

2

=−2F

=2F [along of]

force betweenfo Q and +3q=3F[along of]

net force along of=2F+3F=5F

force between Q and -3q=−3KQq/r

2

=−3F

=3F [along oe]

force betweenfo Q and +q =F[along oe]

net force along of=F+3F=4F

force between Q and -4q =−4KQq/r

2

=−4F

=4F [along od]

force betweenfo Q and +q=F [along od]

net force along of=F+4F=5F

therefore:,

resultant =vector(od)+vector(of)

=5F{along oe}

net resuitant force=5F+4F

=9F[along oe]

therefore n=9

Answer 2

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 velocity that has to be given to -Q such that it escapes from the charge system.

Calculation:

In order to escape the charge system , the -Q has to be provided with the kinetic energy equal to the initial electrostatic potential energy.

Let d be distance of vertex charge from centre.

$\therefore \: KE = PE$

$= > \dfrac{1}{2} m {v}^{2} = 6 \times \bigg \{\dfrac{k(Q \times q)}{ d } \bigg \}$

$= > \dfrac{1}{2} m {v}^{2} = 6 \times \bigg \{\dfrac{kQq}{ a} \bigg \}$

$= > m {v}^{2} = 12\times \bigg \{\dfrac{kQq}{ a} \bigg \}$

$= > {v}^{2} = 12\times \bigg \{\dfrac{kQq }{m a} \bigg \}$

$= > v = \sqrt{12\times \bigg \{\dfrac{kQq }{m a } \bigg \}}$

Putting value of Coulomb's Constant:

$= > v = \sqrt{12\times \bigg \{\dfrac{Qq }{4\pi\epsilon_{0} m a } \bigg \}}$

$= > v = \sqrt{ \dfrac{3Qq }{4\pi\epsilon_{0} m a} }$

So , final answer is :

$\boxed{ \red{ \bold{v = \sqrt{ \dfrac{3Qq }{4\pi\epsilon_{0} m a } }}}}$