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/x^{2}( air )or , 0.1 =9×10

^{9}xq^{2}/(0.5)^{2}or , √0.1×0.25/9×10

^{9}= qor ,q =√0.025/9×10

^{9}=√0.000278×10

^{-8}=√2.78×10

^{-12}∴ q =1.67×10

^{-6 }cIn the second part , a medium of dielectric constant is placed i.e. k =10 , Distance (r) =?

F =1/4π∈

_{o}k .q^{2}/r^{2}or , 0.1 =9×10

^{9}/10 x(1.67×10^{-6})^{2}/r^{2}or , r =√9×10

^{9}x(1.67×10^{-6})^{2}or , r =√25.10×10

^{-3}∴r =0.16 m