Find the magnitude of the resultant force on a charge of `1 muC` held at P due to two charges of `+ 2xx10^(-8)C` at A and B respectively.
Given AP = 10 cm and BP = 5 cm.
`/_APB = 90^(@)`, Fig.

Explanation

Here, `F = ?`
Charge at `P, q = 1muC = 10^(-6)C`
Charge at A, `q_(1) = 2xx10^(-8)C`
Charge at B, `q_(2) = -10^(-8)C`
`AP = 10 cm = 0.1 m, BP = 5 cm =0.05 m`
`/_ APB = 90^(@)`
Force at P due to`q_(1)` to `q_(1)` charge at A,
`F_(1) = (1)/(4pi in_(0)) (q_(1) q)/(AP^(2))`, along AP produced
` = (9xx10^(9)xx2xx10^(-8)xx10^(-6))/((0.1)^(2)) = 18xx10^(-3) N`
Force at P due to `q_(2)` charge at B,
`F_(2) = (1)/(4pi in_(0)) (q_(2) q)/(BP^(2))` , along PB
` = (9xx10^(9)xx10^(-8)xx10^(-6))/((0.05)^(2)) = 36xx10^(-3) N`
An angle between `vec(F_(1))` and `vec(F_(2)) is 90^(@)`,
Resultant force, `F = sqrt((18xx10^(-3))^(2) + (36xx10^(-3))^(2))`
` = 18xx10^(-3)xx2.236`
`= 4.0xx10^(- 2)N`