Consider three charges `q_(1),q_(2)` and `q_(3)` each equal to `q`, at the vertices of an equilateral triangle of side l. What is the force on a charge Q placed at the centroid of the triangle?

Options

A. `(3Qq)/(4piepsilon_(0)l^(2))`
B. `(2Qq)/(4piepsilon_(0)l^(2))`
C. `(Qq)/(2piepsilon_(0)l^(2))`
D. zero

Answer

D

Explanation

As shown in fiqure draw `ADbotBC`.

`therefore" "AD=ABcos30^(@)=(lsqrt3)/2`
Distance AO of the centroid O from A
`=2/3AD=(2l)/3sqrt3/2=l/sqrt3`
`therefore" "` Force on Q at O due to charge `q_(1)` = q at A,
`vecF_(1)=1/(4piepsilon_(0))(Qq)/((l//sqrt3)^(2))=(3Qq)/(4piepsilon_(0)l^(2))`, along AO
Similarly, force on O due to charge `q_(2)` = q at B
`vecF_(2)= (3Qq)/(4piepsilon_(0)l^(2))`, along BO
and force on Q due to charge `q_(3)`=q at C
`vecF_(3)= (3Qq)/(4piepsilon_(0)l^(2))`, along CO
Angle between forces `F_(2) and F_(3) = 120^(@)`
By parallelogram law, resultant of `vecF_(2)andvecF_(3)= (3Qq)/(4piepsilon_(0)l^(2))`, along OA
`therefore" "` Total force on Q = `(3Qq)/(4piepsilon_(0)l^(2))-(3Qq)/(4piepsilon_(0)l^(2))=0`