1.

Two points charges are kept in air with a separation between them. The force between them is F_(1), if half of the space between the charges is filled with a dielectric constant 4 and the force between them is F_(2)." If "(1)/(3)rd of the space between the charges is filled with dielectric of dielectric constant 9. Then (F_(1))/(F_(2)) is

Answer»

`(27)/(64)`
`(16)/(81)`
`(81)/(64)`
`(100)/(81)`

Solution :When dielectric of THICKNESS t is introduced in two charges at distance r, the effective force between the CHARGE is GIVEN by
`F=(q_(1)q_(2))/(4pi epsi_(0)[r-t+tsqrt(K)]^(2))`
where, K = dielectric constant of medium In first case, t=r/2 and k=4
`F=(q_(1)q_(2))/(4pi epsi[r-r//2+(r)/(2)sqrt(4)]^(2))=(q_(1)q_(2))/(4pi epsi_(0)""(9)/(4)r^(2))=(q_(1)q_(2))/(9pi epsi_(0)r^(2))`
In second case, t =r/3 and K=9
`therefore F_(2) =q_(1)q_(2)//4pi epsi_(0)[r-(r)/(3) +(r)/(3)sqrt(3)]^(2)=(q_(1)q_(2))/(4pi epsi_(0)((25)/(9))r^(2))`
`therefore (F_(1))/(F_(2))=(q_(1)q_(2))/(9pi epsi_(0)r^(2))xx(4pi epsi_(0)((25)/(9))r^(2))/(q_(1)q_(2))=(100)/(81)`


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