Question

ello guys !! .. gud Evening !! .. plz help by solving this !! ..​

Answer 1

Electric field due to an electric dipole at a point on its axial line: AB is an electric dipole of two point charges −q and +q separated by small distance 2d. P is a point along the axial line of the dipole at a distance r from the midpoint O of the electric dipole.

The electric fiedl at the point P due to +q placed at B is,

$E_1 = \frac{1}{4\piε0} \frac{q}{ {(r \: - \: d)}^{2} }$

(along BP)

The electric field at the point P due to −q placed at A is,

$E_2 = \frac{1}{4\piε0} \frac{q}{ {(r \: + \: d)}^{2} }$

Therefore, the magnitude of resultant electric field (E) acts in the direction of the vector with a greater, magnitude. The resultant electric field at P is

E= E1 +(−E2)

$E_ = \frac{1}{4\piε0} \frac{q}{ {(r \: - \: d)}^{2} } \: - \: \frac{1}{4\piε0} \frac{q}{ {(r \: + \: d)}^{2} }$

along BP

$E \: = \frac{q}{4\piε0} | \frac{1}{ {(r \: - \: d)}^{2} } \: - \: \frac{1}{ {(r \: + \: d)}^{2} } |$

along BP

$E \: = \frac{q}{4\piε0} | \frac{4rd}{ {( {r}^{2} \: - \: {d}^{2} )}^{2} } \: |$

along BP

$E \: = \frac{q}{4\piε0} - \frac{4rd}{ { {r}^{4} } } \: = \: \frac{q}{4\piε0} \frac{4d}{ { {r}^{3} } } \:$

$E = \frac{1}{4\piε0} \frac{2p}{ { {r}^{3} } } \:$

along BP

[∵ Electric dipole moment p=q×2d]

E acts in the direction of dipole moment.