Three point charges of `+2muC, -3 mu C and -3 muC ` are kept at the vertices A,B and C , respectively of an equilateral triangle of side 20 cm as shown in What should be the sign and magnitude of the charge to be placed at the mid-point (M) of side BC so that the charge at A remains in equilibrium?
` (##U_LIK_SP_PHY_XII_C01_E09_017_Q01.png" width="80%">

Explanation

Charge q on point M should be such that force on A due to it should just nulify the resultant of forces ` oversetto (F_(AB)) and oversetto (F_(AC)) `
Now ` " "F_(AB) =F_(AC) =(1)/(4 pi in _0) .(q_1q_2)/(r^(2)) `
` =(9xx 10^(9) xx 2 xx 10^(-6) xx 3xx 10^(-6))/((0.2)^(2)) =1.35N`
Resultant of `oversetto (F_(AB)) and oversetto (F_(AC)) ` inclined at an angle of `60^(@) ` is
` oversetto F= 2 oversetto (F_(AB)) .cos "" (theta)/(2)=2xx 1.35 xx cos 30^(@) `
` " " = 2xx1.35 xx (sqrt3)/(2)= 1.35 xx sqrt 3N `along AM
For equilibrium force ` oversetto (F_(AM)) ` should be equal and opposite to `oversetto F `
` therefore " " F= |oversetto (F_(AM)) | =(9xx 10^(9) xx 2 xx 10^(-6) xx q)/((AM)^(2)) =(9xx10^(9) xx 2xx 10 ^(-6) xx q)/([(0.2)^(2) -(0.1) ^(2) ])`
` or " " 1.35 xx sqrt3 =(9xx 10^(9) xx2xx 10^(-6) xxq)/(0.03) `
` rArr " " q= (1.35 xx sqrt3 xx 0.03)/(9xx10 ^(9) xx 2 xx 10 ^(-6)) =3.93 xx 10^(-6) C or 3.93 muC `
` (##U_LIK_SP_PHY_XII_C01_E09_017_S01.png" width="80%">