Three charges of equal magnitude q is placed at the vertices of an equilateral triangle of side l. The force on a charge Q placed at the centroid of the triangle is

Explanation

Method 1: The resultant of three equal coplanar vectors acting at a point will be zero if these vectors forn a closed polygon. Hence, the vector sum of the forces `vecF_(1),vecF_(2)`, and `vecF_(3)` is zero.
Method 2: The force action on the charge Q are
`vecF_(1)`= force on Q due to `q_(1)=(1)/(4 pi epsilon_(0)) (Qq_(1))/(AO^(2)) hat(AO)`
`vecF_(2)` = force on Q due to `q_(2)=(1)/(4 pi epsilon_(0)) (Qq_(2))/(BO^(2)) hat(BO)`
,
`vecF_(3)`= force on Q due to `q_(3)=(1)/(4 pi epsilon_(0)) (Qq_(3))/(CO^(2)) hat(CO)`
the resulant force is
`vecF_(R)=vecF_(1)+vecF_(2)+vecF_(3)`
`(1)/(4 pi epsilon_(0)) (Qq)/(AO^(2)) (hat(AO)+hat(BO)+hat(CO))=0`
`(as |q_(1)|= |q_(2)|= |q_(3)|` and `|vec(AO)|=|vec(BO)|=|vec(CO)|`
Also, `vec(AO)+vec(BO)+vec(CO)=0` because these are three equal vectors in a plane making angles of `120^(@)` with each other.
Method 3: The resulatant force `sum vec(F)` is the vector sum of individual force i.e.
`sum vec(F)=vecF_(1)+vecF_(2)+vecF_(3)`
`:. sum F_(x)=F_(1x)+F_(2x)+F_(3x)`
`=0+F_(2) cos 30^(@)` ...(i)
and `sum F_(y)=F_(1y)+F_(2y)+F_(3y)`
`=-F_(1)+F_(2) sin 30^(@)+F_(3) sin 30^(@)` ...(ii)
As `|F_(1)|=|F_(2)|=|F_(3)|=|F|` (say) Eqs.(i) and (ii) become
`sumF_(x)=0` and `sumF_(y)=0`. Hence resultant force `sumvec F=0`