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,

Answer 1

Answer:

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

Explanation:

Electric potential energy at the centre due to any single charge q and -Q

$U = -\frac{kQq}{a}$

Net electric potential at the centre = 6U (as there are six identical charges 'q')

$U' = 6U = \frac{-6KQq}{a}$

When the charge -Q is taken far away from the system (say, infinite distance), the potential energy on the charge becomes 0.

Hence, due to conservation of energy, we can write,

U'' - U' = K.E

U'' = energy far away from the system

K.E = Kinetic energy

$-(-\frac{6kQq}{a}) = \frac{mv^2}{2}$

$v^2 = \frac{12kQq}{ma}\\\\v = \sqrt{\frac{12kQq}{ma}}$

This should be the required velocity.