Three charges are placed at the vertices of an equilateral trianlge of side `a` as shown in the following figure. The force experienced by the charge placed at the vertex `A` in a direction normal to `BC` is

Options

A. `(Q^(2))/(4piepsilon_(0)a^(2))`
B. `-(Q^(2))/(4piepsilon_(0)a^(2))`
C. zero
D. `(Q^(2))/(2piepsilon_(0)a^(2))`

Answer

c

Explanation

Resultant force
`F.=sqrt(F^(2)+F^(2)+2FFcos120^(@))=F[becausecos120^(@)=(-1)/(2)]`

Now, from figure
`F=sqrt(F.^(2)+F..^(2)+2F.F..cos90^(@))`
Now, the force normal to BC at vertex A is
`F.=sqrt(F^(2)-F.^(2))=0 " " (becauseF.=F)`