Two charges `q_1 and q_2` are placed on the X-axis at point x=-a and x =+a respectively.
a. How must `q_1 and q_2` be related for the net electrostatic force on a charge Q at x=`+(a)/(2) ` to be zero?
b. Repeat (a) but with +Q at x =+ ` (3a)/( 2) `

Explanation

Data supplied
` (##JYT_AJP_AIO_PHY_XII_C01_SLV_014_S01.png" width="80%">
Net force at Q is zero
Here force due to `q_1` acts towards left and that due to` q_2` towards right . Hence the resultant force on Q is `F_(Qq_1) -F_(Qq_2) =0`
i.e., `F_(Qq_1) = F_(Qq_2) `
`(1)/( 4pi in _0) (Qq_1)/( (a+(a)/(2)))=(1)/( 4pi in _0) .(Qq_2)/( (a)/(2)^(2)) =( q_1)/(( 9a^(2))/( 4)) =(q^(2))/( (a^(2))/( 4)) `
` (q_1)/( 9) =q_2" " therefore q_1 =9q_2`
b. ` (##JYT_AJP_AIO_PHY_XII_C01_SLV_014_S02.png" width="80%">
Here force due to `q_1 and q_2` both act towards right. Hence the resultant force on Q is `F_(Qq_1) +F_(Qq_2) =0`
` F_(Qq_1) =-F_(Qq_2) `
` (1)/( 4pi in _0) .( Qq_1)/( (a+(3a)/(2))^(2)) =(1)/( 4pi in _0) .(Qq_2)/( (a)/(2))^(2) =( q_1)/((25a^(2))/( 4)) =(-q_2)/( (a^(2))/( 4))`
` (q,)/( 25) =q_2 " " q_1 =-25 q_2`