Three point charges of `+2muC, -3muC` , and `-3muC` are kept at the vertices A, B, and C, respectively of an equilatreral 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

`(1.5 sqrt(3))/(3)xx10^(-12)=(sqrt(3))/(2)pC`, positive charge

Explanation

Net electric field at A due to charges at B and C is
`E_(A)=2E_(AC) sin 60^(@)=2xx9xx10^(9)xx(3xx10^(-6))/((0.20)^(2))xx(sqrt(3))/(2)`
`sqrt(3)xx6.75xx10^(5)`
`AM=sqrt((20^2)-(10^2))=sqrt(400-100)=10sqrt(3) cm`

Let the charges at M be q. Charge q should be positve so that there can be repulsion between the charges at A and M.
`E_(AM)=9xx10^(9)xx(q)/((10sqrt(3)//100)^(2))=3qxx10^(11)`
For A to be in equilibrium
`E_(A)=E_(AM) implies sqrt(3)xx6.75xx10^(5)=3qxx10^(11)`
or `q=(9sqrt(3))/(4)xx10^(-6)=(9sqrt(3))/(4)muC`.