Two equal positive charges, each of `2muC` interact with a third positive charge of `3 muC` situated as shown in Fig. Calculate the magnitude and direction of the force on the ` 3 muC` charge.

Explanation

In fig ,
`OA = OB = 3 m, OP = 4 m`
`:. AP = BP = sqrt(3^(2) + 4^(2)) = 5m`
According to Coulomb's law,
Force on charge at P due to charge at A
`F_(1) = (1)/(4pi in_(0)) (q_(1) q_(2))/((AP)^(2))`
`F_(1) = (9xx10^(9)xx(2xx10^(-6))xx(3xx10^(-6)))/(5^(2))`
`= (54)/(25) xx 10^(-3) N = 2.16xx10^(-3) N`
`F_(1) = 2.16XX10^(-3) N`, along `PA'` opposite to `PA`. It has two recentangular components `F_(2) cos 0` along ` PX` and `F_(1)` sin 0 along `PY'`. Similarly, force on charge along `PB` opposite to `PB'`. It also has two recentangular components :` F_(2) cos 0` along `PX` and `F_(2) sin 0` along `PY`.
The components along `PY'` cancel. The components along `PX` add up.
Total force on `3muC` charge is `F = 2 F_(1) cos theta`
` = 2xx2.16xx10xx10^(-3)xx(4)/(5)`
` = 3.5xx10^(-3) N`, along PX.