Three point charges of `+2muC, -3muC` , and `-3muC` are kept at the vertices A, B, and C, respectively of an equilateral triangle of side 20 cm. what should be the sign and magnitude of the charge (q) to be placed at the midpoint (M) of side BC so that the charge at A remains in equilibrium?

Answer

`6.49 muC`, positive

Explanation

as shown in Fig, force exertend on charege `+3muC` by charge at B
`F_(1) = (1)/(4pi in_(0)) (q_(1) q_(2))/(r^(2)) = (9xx10^(9)xx3xx10^(-6)xx5xx10^(-6))/((0.2)^(2))`
`= 3.375 N` along AB
Force exerted on `+3 muC` charge by charge at C
`F_(2) = (9xx10^(9)xx3xx10^(-6)xx5xx10^(-6))/((0.2)^(2))`
`= 3.375 N` along Ac
Resultant force of `F_(1) and F_(2)`
`F = sqrt( F_(1)^(2) + F_(2)^(2) + 2 F_(1) F_(2) cos 60^(@)) = 5.845 N`
For charge A to be in equlibrium positive charge must be placed at M, to exert force along MA
`AM = sqrt((0.2)^(2) - (0.1)^(2)) = 0.1xx sqrt(3) m`
Force on charge `+3 muC` will be zero if
`(9xx10^(9) xx q xx3xx10^(-6))/((0.1xx sqrt(3))^(2)) = 5.845`
`q = 6.49 muC`