Physics Electric Charges and Fields

Electric field due to a dipole : Electric field at axis (Line joining the charges)

Electric field due to a short dipole on its axis at a point A at a distance r from dipole (l < < r ) :

`E_A = q/((4pi epsilon_0)(r-l)^2) - q/((4pi epsilon_0)(r+l)^2)`

`E_A = (4qlr)/((4pi epsilon_0)(r^2-l^2)^2`

after using dipole approximation ` {l<
`vecE_A = (2vecP)/((4piepsilon_0)r^3)`

Electric field at equator: (Line perpendicular to axis passing through centre)

Electric field at a point distance rfrom the centre of the short dipole ( l < < r)

`E_B = 2[{q/((4piepsilon_0)(r^2+l^2)}costheta]`

`E_B = (2q)/((4piepsilon_0)(r+l)^2) l/sqrt(l^2 + r^2) = (2ql)/((4piepsilon_0)(r^2+l^2)^(3/2)`

after using dipole approximation `l < < r`

`vecE_B = -vecP / (4piepsilon_0 r^3)` (-ve sign indicate that the field is oppositly directed to dipole direction)

Electric field at any point `A (r, theta)` due to dipole

LetA be a point at a distance r from the mid-point 0 of the dipole. Let `theta` be the angle between OA and
the dipole moment p. Since dipole moment is a vector so we can resolve its components `pcostheta` and `p sintheta`
along and perpendicular to OA. Due to pcose (axial point) electric fie ld wi ll be in the direction of `pcostheta`
and due to `p sintheta` (equatorial position) electric field wi ll be opposite to `p sintheta`

Electric field at A

`E = sqrt((2kp costheta/r^3)^2 + (kp sintheta/r^3)^2) = (kp/r^3)sqrt(1+3costheta^2)`

& `tanalpha = E_theta / E_r = sintheta/(2costheta) = tantheta/2`