In a certain co-ordinate system, charge, `q_(1) = -2xx10^(-4) C` is at `x = 0, y = 0`, charge `q_(2) = 1xx10^(-3)C` is at `x = 10 m` and `y = 0` and charge `q_(3) = 1xx10^(-4)C` is at `x = 0, y = 5cm`, Find the magnitude of resultantant force on `q_(1)`

Answer

`19.4 N`

Explanation

Refer to fig
`F_(1)` is force on `q_(1)` due to `q_(2)`
`F_(1) = (9xx10^(9)xx2xx10^(-4)xx10^(-3))/((10)^(2)) = 18N`
`F_(2)` is force on `q_(1)` due to `q_(3)`
`F_(2) = (9xx10^(9)xx2xx10^(-4)xx1xx10^(-4))/((5)^(2)) = 7.2N`
As `F_(1) and F_(2)` are perpendicular to each other,
resultant force, `F = sqrt(F_(1)^(2) + F_(2)^(2))`
`= sqrt((18)^(2) + (7.2)^(2)) = 19.4 N`