find the expression for electric field of a dipole when it is in end of position and justify the answer and hence find expression for ideal gas. Please give me 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
If the point P is far away from the dipole, then d≪r
Answers & Comments
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
If the point P is far away from the dipole, then d≪r
∴E=
4πε
0
q
−
r
4
4rd
=
4πε
0
q
r
3
4d
E=
4πε
0
1
r
3
2p
along BP