Charges `q_(1) = 1.5 mC, q_(2) = 0.2 mC and q_(3) = -0.5 mC`, are placed at points A,B,C respectively as shown in Fig. If `r_(1) = 1.2m and r_(2) = 0.6m`, calculatae magnitude of resultant force on `q_(2)`.

Answer

`3.1xx10^(3)N`

Explanation

Refer to Fig.
`F_(1) = (q_(1) q_(2))/(4pi in_(0) r_(1)^(2))`
`= (1.5xx10^(-3)xx9xx10^(9))/((1.2)^(2))`
`= 1.875xx10^(3) N` …. along AB produced
`F_(2) = (q_(2) q_(3))/(4pi in_(0) r_(2)^(2)) = (0.2xx10^(-3)xx9xx10^(9))/((0.6)^(2))`
`= 2.5xx10^(3) N` along `BC _|_ AB`.
Resulant force on `q_(2) = sqrt(F_(1)^(2) + F_(2)^(2))`
`= 3.1xx10^(3)N`