E vector axis =2kp/r^3 derive this equation more of electric feild at point p
Answers
Answer:
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
=
4πε
0
1
(r−d)
2
q
(along BP)
The electric field at the point P due to −q placed at A is,
E
2
=
4πε
0
1
(r+d)
2
q
(along PA)
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=E
1
+(−E
2
)
E=[
4πε
0
1
(r−d)
2
q
−
4πε
0
1
(r+d)
2
q
] along BP
E=
4πε
0
q
[
(r−d)
2
1
−
(r+d)
2
1
] along BP
E=
4πε
0
q
[
(r
2
−d
2
)
2
4rd
] along BP