An electric dipole is prepared by taking two electric charges of `2 xx 10^(-8)` C separated by distance 2 mm. This dipole is kept near a line charge distribution having density `4 xx 10^(-4)` C/m in such a way that the negative electric charge of the dipole is at a distance 2 cm from the wire as shown in the figure. Calculate the force acting on the dipole.
[Take `k=1/(4piepsilon_(0)) = 9 xx 10^(9) Nm^(2)C^(-2)`]

Explanation

`k = 9 xx 10^(9) Nm^(2)C^(2)`
`lambda = 4 xx 10^(-4) C//m`
`q = 2 xx 10^(-8)` C
`r_(+) = 2.2 cm = 2.2 xx 10^(-2)` m
`r_(infty) = 2 cm = 2 xx 10^(-2)` m
The electric field intensity at some point r from continuous line charge distribution having density X is given by the formula,
`E = lambda/(2piepsilon_(0)).l/r = (2klambda)/r`
`vecF = (-2klambdaqhati)/r_(-)` and `vecF_(+) = (2klambdaqhati)/r_(+) (therefore F = Eq)`
`therefore` Resultant force,
`vecF = vecF_(+) + vecF_(-) = 2klambda q[1/r_(+) - 1/r_(-)]hati`
`therefore vecF = 2 xx 9 xx 10^(9) xx 4 xx 10^(-4) xx 2 xx 10^(-6)`
`vecF = 144 xx 10^(-1) [-0.04545] hati`
`therefore vecF = -0.6545 hati N`