Three point charges `q,-2q` and q are located along the x-axis a s shown in figure. Show that the electric field at `P (y gtgta)` along the y-axis is,
`E=-1/(4piepsilon_0) (3qa^2)/y^4 hatj`

Answer

A

Explanation

`E_1=` Electric field at `P` due to `q`
`=(1/(4piepsilon_0))q/r^2`
`E_2=` Electric field at `P` due to `-2q`
`=(1/(4piepsilon_0))(2q)/y^2`
`:.` Net electric field at P
`E=2E_1 costheta+E_2`
`=2(1/(4piepsilon_0))(q/r^2)(y/r)+(1/(4piepsilon_0))((2q)/y^2)`
`=(2q)/(4piepsilon_0)[y/r^3]+[1/y^2]`
`=(2q)/(4piepsilon_0)[y/(a^2+y^2)3//2+1/y^2]`
`=(2q)/(4piepsilon_(0))[y/y^(3)+(1+a^2/y^2)^(-3//2)+1/y^2]`
Applying binomial expression for `a^2/y^2 lt lt 1` we get, `E=(2q)/(4epsilon_(0))(-3/2a^2/y^4)`
`=-(3qa^(2))/(4piepsilon_0y^4)`
`E=-(3qa^(2))/(4piepsilon_0y^4)hatj`