A the third body which is uncharged contact with body A then both distribute net charge equally. e.g. Qnet=(0+qa) because they equally distribute so, both have finally qa/2 charge available. now third body contact with body B then Qnet=(qa/2+qb) now both distribute charge equally hence, each has (qaRead more
A the third body which is uncharged contact with body A then both distribute net charge equally.
e.g. Qnet=(0+qa)
because they equally distribute so, both have finally qa/2 charge available.
now third body contact with body B
then Qnet=(qa/2+qb)
now both distribute charge equally
hence,
each has (qa/2+qb)/2
hence finally charge in body A=qa/2
charge in body B=(qa/2+qb)/2=qa/4+qb/2
Given, Force (f) =0.1 N Distance between Charge (r) =540 cm Charge (q) =? we know that , F = 1/4π∈o.qxq/x2 ( air ) or , 0.1 =9x109xq2/(0.5)2 or , √0.1x0.25/9x109 = q or ,q =√0.025/9x109 =√0.000278x10-8 =√2.78x10-12 ∴ q =1.67x10-6 c In the second part , a medium of dielectric constanRead more
Given,
See lessForce (f) =0.1 N
Distance between Charge (r) =540 cm
Charge (q) =?
we know that ,
F = 1/4π∈o.qxq/x2 ( air )
or , 0.1 =9×109xq2/(0.5)2
or , √0.1×0.25/9×109 = q
or ,q =√0.025/9×109
=√0.000278×10-8
=√2.78×10-12
∴ q =1.67×10-6 c
In the second part , a medium of dielectric constant is placed i.e. k =10 , Distance (r) =?
F =1/4π∈ok .q2/r2
or , 0.1 =9×109/10 x(1.67×10-6)2/r2
or , r =√9×109x(1.67×10-6)2
or , r =√25.10×10-3
∴r =0.16 m