Zero error of an instrument introduces

Answer

`3.17xx10^(-9) C`

Explanation

`F_(B)=2F cos 30^(@)=(2Kq^(2))/a^(2)xxsqrt(3)/2=(sqrt(3)Kq^(2))/a^(2)`
For Charge at B: `2r cos 30^(@)=a`

`r=a/sqrt(3)`
`T sin theta=F_(B)`
`T cos theta=mg`
`tan theta=F_(B)/(mg)`
`rArr (sin theta)/(cos theta)=(99xx10^(-2))/(sqrt(3)xx10^(-2))=(sqrt(3)xx9xx10^(9)xxq^(2))/((3xx10^(-2))2)xx1/(10^(-3)xx10)`
`rArr q=3.17xx10^(-9)C`