Two charges each equal to `eta q(eta^(-1) lt sqrt(3))` are placed at the corners of an equilateral triangle of side `a`. The electric field at the third corner is `E_(3)` where`(E_(0)= q//4piepsilon_(0)a^(2))`

Options

A. `E_(3)=E_(0)`
B. `E_(3) lt E_(0)`
C. `E_(3) gt E_(0)`
D. `E_(3) ge E_(0)`

Answer

c

Explanation

`E_(1)= (etaq)/(4piepsilon_(0)a^(2))`, `E_(2)=(eta q)/(4piepsilon_(0)a^(2))`
Therefore `E = |vec(E_(1)) +vec(E_(2))|`
`=sqrt(E_(1)^(2)+E_(2)^(2)+2E_(1)E_(2) cos 60^(@))=(sqrt(3)etaq)/(4piepsilon_(0)a^(2))`. Since `eta^(-1) lt sqrt(3),1 lt sqrt(3)eta, sqrt(3 eta) gt 1`
`implies (sqrt(3)etaq)/(4piepsilon_(0)a^(2)) gt (q)/(4piepsilon_(0)a^(2))rArr E_(3) gt E_(0)(E_(0)= (q)/(4piepsilon_(0)a^(2)))`.